Category: CBSE Class 10

On this page, you will find Statistics Class 10 Notes Maths Chapter 14 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 14 Statistics will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Maths Chapter 14 Notes Statistics

Statistics Class 10 Notes Understanding the Lesson

The measures of central tendency are:

• Arithmetic mean or mean
• Median
• Mode

Mean of Raw Data

Mean of n observations x_1; x₂, x₃, ..x_n is given by
\(\bar{x}=\frac{x_{1}+x_{2}+x_{3}+\ldots+x_{n}}{n}=\frac{\Sigma x_{i}}{n}\)
where ∑ (sigma) means “summation of’.

1. Mean of Grouped Data

(i) Direct method:
\(\bar{x} \doteq \frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\)
For each class interval, Class mark = \(\frac{\text { Lower limit }+\text { Upper limit }}{2}\)

(ii) Short-cut method or assumed mean method:

Where a = assumed mean

(iii) Step-deviation method:

h=Class-Size

2. Mode of Groped Data
Class with the maximum frequency is called the modal class.
\(\text { Mode }=l+\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right] \times h\)

Where l = lower limit of the modal class
h = size of the class-interval
f₁ = frequency of the modal class
f₀ – frequency of the class preceeding the modal class
f₂ = frequency of the class succeeding the modal class

3. Median of Ungrouped Data
To find the median of ungrouped data, first arrange the data values of the observations in the ascending or descending order. Then,

4. Median of Grouped Data
\(\text { Median }=l+\left[\frac{\frac{n}{2}-c \cdot f \cdot}{f}\right] \times h\)
l – lower limit of median class.
n = number of observation
c.f. = cumulative frequency of the class preceeding the median class
f = frequency of the median class
h = class size

Median class: Class whose cumulative frequency is greater than (and nearest to) \(\frac{n}{2}\)

On this page, you will find Metals and Non-metals Class 10 Notes Science Chapter 3 Pdf free download. CBSE NCERT Class 10 Science Notes Chapter 3 Metals and Non-metals will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Science Chapter 3 Notes Metals and Non-metals

Metals and Non-metals Class 10 Notes Understanding the Lesson

1. Element is the simplest form of matter which contains one kind of atoms. About 118 elements are known today. There are more than 90 metals, 22 non-metals and a few metalloids.

• Sodium (Na), potassium (K), magnesium (Mg), aluminium (Al), calcium (Ca), Iron (Fe), Barium (Ba) are some metals.
• Oxygen (O), hydrogen (H), nitrogen (N), sulphur (S), phosphorus (P), fluorine (F), chlorine (Cl), bromine (Br), iodine (I) are some non-metals.

2. Physical Properties of Metals

• Metals in their pure state, have a shining surface. This property is called metallic lustre.
• Metals can be beaten into thin sheets. This property is called Gold and silver are the most malleable metals.
• Metals have ability to be drawn into thin wires. This property is called ductility. Gold is the most ductile metal.
• Metals are good conductors of heat and have high melting points. The best conductor of heat are silver and copper. Lead and mercury are comparatively poor conductors of heat.
• The metals that produce a sound on striking a hard surface are said to be
• Alkali metal (Li, Na, K) are so soft that they can be cut with a knife. They have low densities and low melting points.
• Metals have high melting point but gallium and caesium have very low melting points. These two metals will melt if you keep them on your palm.

3. Physical Properties of Non-Metals

• Non-metal are either solids or gases except bromine which is a liquid.

4. Some other exceptions:

• Iodine is a non-metal but it is lustrous.
• Carbon is a non-metal which exist in two allotropic forms: diamond and graphite. Diamond is the hardest substance with a very high melting point. Graphite is a conductor of electricity.

5. Chemical Properties of Metals

I. All metals combine with oxygen to form metal oxides.
Metal + Oxygen → Metal oxide
For example,
2Cu + O₂ → 2CuO
4Al + 3O₂ → 2Al₂O₃

6. Remember
Different metals exhibit different reactivity towards oxygen. Metals such as K and Na react so vigorously that they catch fire if kept in the open. Hence, to protect them they are kept immersed in kerosene oil.

Generally metal oxides are basic in nature. But some oxides like aluminium oxide, zinc oxide show both acidic as well as basic behaviour.

Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

Most metal oxides are insoluble in water but some of these dissolve in water to form alkalies. Sodium oxide and potassium oxide dissolve in water to produce alkali as follows:

Na₂O(s) + H₂O(l) → 2NaOH(aq)
K₂O(s) + H₂O(l) → 2KOH(aq)

II. Metals react with water and produce a metal oxide and hydrogen gas. Metal oxides that are soluble in water dissolve in it further to form metal hydroxide. But all metals do not react with water.

Metal + Water → Metal Oxide + Hydrogen
Metal Oxide + Water → Metal hydroxide

Potassium and sodium react violently with cold water. The reaction is exothermic, so the released hydrogen catches fire immediately.

2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g) + Heat
2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g) + Heat

The reaction of calcium with water is less violent. The heat evolved is not sufficient for the hydrogen to catch fire.
Ca(s) + 2H₂O(l) → Ca(OH)₂(aq) + H₂(g)
Mg(s) + 2H₂O(l) → Mg(OH)₂(aq) + H₂(g)

• Calcium starts floating because the bubbles of hydrogen gas formed stick to the surface of the metal.
• Magnesium does not react with cold water. It reacts with hot water to form magnesium hydroxide and hydrogen.

Aluminium, iron and zinc do not react either with cold or hot water. But they react with steam to form the metal oxide and hydrogen.

2Al(s) + 3H₂O(g) → Al₂O₃(s) + 3H₂(g)
3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)

Lead, copper, silver and gold do not react with water at all.

III. Metals react with acids to give a salt and hydrogen gas.
Metal + Dilute acid → Salt + Hydrogen

• Hydrogen gas is not evolved when a metal reacts with HNO₃. It is because HNO₃ is a strong oxidising agent. It oxidises the H₂ produced to water and itself get reduced to any of the nitrogen oxides.
• Magnesium (Mg) and manganese (Mn) react with very dilute HNO₃ to evolve H₂
• Copper does not react with dilute HCl.

IV. Reactive metals can displace less reactive metals from their compounds in solution or molten form.

Metal A + Salt solution of B → Salt solution of A + Metal B

Fe + CuSO₄ → FeSO₄ + Cu
Zn + CuSO₄ → ZnSO₄ + Cu
Reactivity of metal can be explained on the basis of displacement reactions

7. Knowledge Plus
Aqua regia (Latin for ‘royal water) is a freshly prepared mixture of concentrated HCl and concentrated nitric acid in the ratio of 3:1. It is a highly corrosive, fuming liquid and dissolve gold and platinum.

8. The Reactivity Series
The reactivity series is a list of metals arranged in the order of their decreasing activities.

Metals occupying higher position in the series have more tendency to lose electrons and are more reactive. The metals at the bottom of the series are least reactive. Thus, potassium is the most reactive metal.

9. How do metals and non-metals react?

Reactivity is the tendency of elements (metals and non-metals) to attain a completely filled valence shell. Metal atoms having 1, 2 or 3 electrons in their outermost shell can lose electrons to non-metal atoms having 5,6 or 7 electrons to attain the electronic configuratìoñ of the nearest noble gas (i.e., completely filled valence shell).

Thus the metal atom becomes a positively charged ion or cation and the non-metal atom becomes a negatively charged ion or anion. The cation and anion being oppositely charged attract each other and are held by strong electrostatic forces of attraction to exist as an ionic compound. For example,

Metal ore heated strongly in limited or no supply of air (Calcination).

10. Reduction of Metal Oxide:

(i) Using coke: Coke as a reducing agent.

(ii) Using displacement reaction: Highly reactive metals like Na, Ca and A1 are used to displace metals of lower reactivity from their compounds. These displacement reactions are highly exothermic.

11. Thermite Reaction: Reduction of a metal oxide to form metal by using aluminium powder as a reducing agent. This process is used to join broken pieces of heavy iron objects or welding.

12. Extracting Metals at the Top of the Activity Series:

• These metals have more affinity for oxygen than carbon so they cannot be obtained from their compounds by reducing with carbon.
• They are obtained by electrolytic reduction, for example, Sodium is obtained by electrolysis of its molten chloride                     NaCl→ Na⁺ + Cl^–

As electricity is passed through the solution, metal gets deposited at the cathode and non-metal at the anode.

• At cathode: Na⁺ + e^– →Na
• At anode: 2Cl^– →Cl₂ (g) + 2e^–

III. Refining of Metals

• Impurities present in the obtained metal can be removed by electrolytic refining.

Copper is obtained using this method. Following are present inside the electrolytic tank.

• Anode – slab of impure copper
• Cathode – slab of pure copper
• Solution – aqueous solution of copper sulphate with some dilute sulphuric acid.
• From anode, copper ions are released in the solution and equivalent amount of copper from solution is deposited at cathode.
• Insoluble impurities containing silver and gold gets deposited at the bottom of anode as anode mud.

13. Corrosion

• Metals are attacked by substances in the surroundings like moisture and acids.
• Silver—It reacts with sulphur in air to form a black coating of silver sulphide.
• Copper—It reacts with moist carbon dioxide in air and forms a green coating of copper carbonate.
• Iron—acquires a coating of a brown flaky substance called rust. Both air and moisture are necessary for rusting of iron. Rust is hydrated Iron (III) oxide e., Fe₂O₃.xH₂O

14. Prevention of Corrosion

• Rusting of iron is prevented by painting, oiling, greasing, galvanizing, chrome plating, anodising and making alloys.
• In galvanization, iron or steel is coated with a thin layer of zinc. Zinc oxide formed due to oxidation is impervious to air and moisture protecting further layers from corrosion.

15. Alloys: These are homogeneous mixture of metals with metals or non-metals.
Adding small amount of carbon makes iron hard and strong.

Name of Alloy	Properties	Constituent metal/ Non-metal
1. Steel	Hard	Iron and carbon
2. Stainless steel	Hard, rust free	Iron, nickel and chromium
3. Brass	Low electrical conductivity than pure metal	Copper and zinc
4. Bronze	Hard and easily cast	Copper and tin
5. Solder	Low MP, used to weld wires	Lead and tin
6. Amalgam	Used by dentists	Mercury and any other metal

Class 10 Science Chapter 3 Notes Important Terms

Corrosion is the eating up of metals by the action of air, moisture or a chemical on their surface.

Rust is mainly hydrated iron (III) oxide Fe₂O₃.xH₂O due to corrosion.

Ores are the minerals from which metals can be extracted conveniently and profitably.

Minerals are natural materials in which the metals or their compounds are found in the Earth.

Covalent bond is the chemical bond formed by sharing of electrons between two atoms. Aqua-regia is a freshly prepared mixture of 1 part of concentrated nitric acid and 3 parts of concentrated hydrochloric acid.

Brass is an alloy of copper and zinc. Bronze is an alloy of copper and tin.

Metallurgy is the process of extraction of a metal from its ore and its refining.

Activity series is the arrangement of metals in the order of decreasing reactivity.

On this page, you will find Surface Areas and Volumes Class 10 Notes Maths Chapter 13 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 13 Surface Areas and Volumes will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Maths Chapter 13 Notes Surface Areas and Volumes

Surface Areas and Volumes Class 10 Notes Understanding the Lesson

Surface area: Surface area of an object is the measure of the total area that the surface of an object occupies.

Volume: Volume of an object is the measure of space occupied by the object.

Basic Solids: In standard X, we have studied the surface area and volume of solids. Here we will study more about them.

1. Cuboid
(i) Surface area of cuboid = 2(lb + bh + lh) sq.unit
where l is the length
b is the breadth
h is the height

(ii) Area of four walls of cuboid
= 2(l + b) x h
= [Perimeter of floor x Height] sq. unit

(iii) Surface area of cuboid without roof or lid
= lb + 2 [bh + Ih] sq. unit

(iv) Volume of cuboid = l x b x h unit

(v) Diagonal of cuboid or length of longest rod kept =\(\sqrt{l^{2}+b^{2}+h^{2}}\)unit

2. Cube
Let each edge of a cube be of length a unit. Then
(i) Surface area of cube = 6 side² = 6a² unit

(ii) Surface area of four walls of cube = 4 side²
= 4a² sq. unit

(iii) Surface area of a cube without lid (or rod) of a cube
= 5a² sq. unit.

(iv) Length of longest diagonal (or rod) of a cube
\(=\sqrt{a^{2}+a^{2}+a^{2}}=\sqrt{3}\) aunit

(v) Volume of cube = a³ unit

3. Cylinder

(i) Curved surface area of cylinder = 2πr x h
= Perimeter of base x height sq. unit

(ii) Total surface area of cylinder
= CSA + Area of 2 circular ends of cylinder
= 2πrh + 2πr² = 2πr (r + h)

(iii) Volume of cylinder =πr²h

(iv) Volume of material in hollow pipe = Exterior volume – Interior volume
= πR²h – πr²h = πh [R² – r²]

(v) Total surface area of hollow cylinder
= CSA of outer and inner cylinder + 2(area of base ring)
= 2πRh + 2πrh + 2(πR² – πr²)
= 2π(R + r)h + 2π(R² – r²) = 2π(R + r) (h + R – r)

Note:

• Two ends of cylinder are circles having each area = πr²
• Mass of cylinder = Volume of cylinder x Density
  M = V x ρ

4. Cone

h – OA = height of cone
r = OB = radius of cone
l = AB = slant height of cone

(i) \(l=\sqrt{r^{2}+h^{2}}\) units

(ii) Curved surface area of cone or lateral
surface area of cone = πrl sq. unit

(iii) Total surface area of cone = CSA + Area of circular base
= πrl + πr² – πr(r + l) sq. unit

(iv) Volume of cone =\( \frac{1}{3}\) πr²h cu.unit

5. Sphere

• Surface area of sphere = 4πr² unit
• Volume of sphere = \( \frac{4}{3}\)cu.unit

6. Hemisphere

• Curved surface area of hemisphere = 4πr² sq unit
• Volume of hemisphere = \( \frac{2}{3}\) πr² cu unit
• Total surface area of hemisphere = 2πr² + πr² = 3πr² sq. unit

7. Spherical shell

(i) Total surface area of spherical shell = 4πR² + 4πr²
= 4πr(R² + r²) sq. unit

(ii) Volume of spherical shell = \( \frac{4}{3}\)π(R³– r³) cu . unit

Shapes of Frustum

(i) Slant height of frustum = \(\sqrt{(\mathrm{R}-r)^{2}+h^{2}}\) unit
(ii) Curved surface area of frustum = π(R + r)l sq. unit
(iii) Total surface area of frustum of cone
= πl (R + r) + πR² + πr² sq. unit

(iv) Volume of frustum of cone = \(\frac{1}{3}\)πh (R² + r² + Rr) sq. unit

Volume of Combination Solids

The volume of the solid formed by joining two basic solids will actually be the sum of the volumes of the two basic solids.

Conversion of Solid from One Shape to Another

If we melt the candle in the shape of cylinder and pour it into a conical vessel, then it changes into the conical shape. Thus, volume of cylindrical candle = Volume of conical solid.

On this page, you will find Acids Bases and Salts Class 10 Notes Science Chapter 2 Pdf free download. CBSE NCERT Class 10 Science Notes Chapter 2 Acids Bases and Salts will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Science Chapter 2 Notes Acids Bases and Salts

Acids Bases and Salts Class 10 Notes Understanding the Lesson

1. Acids and Bases: Acids are sour in taste and change the colour of blue litmus to red. The term has been derived from the Latin word ‘acidus’ which means sour taste. Generally acids have atleast one or more hydrogen atoms in their formulae.

An acid may be defined as a chemical substance which releases one or more H⁺ or HsO⁺ ions in aqueous solution.
For example, HCl, HNO₃, H₂SO₄, etc.

Bases are bitter and change the colour of the red litmus to blue. Generally bases have one or more hydroxyl (OH^–) groups. They produce hydroxyl ions (OH^–) when dissolved in water.

A base may be defined as a chemical substance which releases one or more OH^– ions in aqueous solution.
For example, NaOH, KOH, etc.

• Acid-Base indicator: Natural/synthetic materials which indicate the presence of acid or base in a solution, are called acid base indicator or simply indicator.
• Litmus solution: It is a purple dye which is extracted from lichen, a plant belonging to the division
  .
• Phenolphthalein: It is a colourless organic dye in acidic or neutral medium but it changes to pink in basic medium.
• Methyl orange: It is an orange coloured dye and keeps this colour in the neutral medium. In the acidic medium, the colour of the indicator becomes red and in the basic medium, it changes to yellow.
• Red cabbage juice: Its colour remains red in acidic medium but changes to green if the medium is basic or alkaline.
• Turmeric solution: It is a yellow dye and in the acidic as well as neutral medium, its colour remains yellow. In the basic medium the colour changes to reddish brown.

2. Olfactory indicators: These are chemical substances whose odour changes in acidic or basic medium.
For example, onion, vanilla and clove oil.

3. Reaction of acid or base with metal: Metals react with acids to liberate hydrogen gas and form salt.
Acid + Metal → Salt + Hydrogen gas

A few metals like zinc, lead and aluminium react with bases to give off hydrogen.

4. Reaction of acids with metal hydrogen carbonate and metal carbonates: All metal carbonates and hydrogen carbonates react with acids to give the corresponding salt, carbon dioxide and water.

Metal carbonate + Acid → Salt + H₂O + CO₂
Metal hydrogen carbonate + Acid → Salt + H₂O + CO₂

For example,
Na₂CO₃(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)
NaHCO₃(s) + HCl(ag) → NaCl(aq) + H₂O(Z) + CO₂(g)

The released CO₂ gas turns lime water milky due to formation of CaCO₃.

On passing excess CO₂, the milky white precipitate dissolves in water.

5. Neutralisation reaction: A chemical reaction between an acid and a base to give a salt and water is known as neutralisation reaction. In general neutralisation reaction can be written as

Base + Acid → Salt + Water
NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

(i) Reactions of metal oxides with acids

(ii) Reactions of non-metallic oxide with base
Non-metallic oxide + Base→ Salt + Water
CO₂ + Ca(OH)₂ → CaCO₃ + H₂O

How strong are acid or base solutions: Any aqueous solution, be it acidic, alkaline or neutral, will have both H⁺ and OH^– ions.
The  solution will be acidic or alkaline depending upon which of the two ions is present in larger concentration.

A scale for measuring hydrogen ion concentration in a solution, called pH scale has been developed. pH scale was given by a Danish chemist Sorensen. The p in pH stands for ‘potenz’ in German meaning ‘power’.

pH should be thought of simply as a number between 0-14 which indicates the acidic or basic nature of a solution. Higher the hydrogen ion concentration, lower is the pH value.

Knowledge +
pH = negative logarithm to the base 10 of the H⁺ ion concentration.
The concentration is in mol/dm³. pH = – log H⁺

Remember: Every one fold change in the pH scale brings about a ten-fold change in H⁺ ion concentration. The strength of acids and bases depends on the concentration of H⁺ ion and OH^–.

If we take two acids HCl and CH₃COOH of the same concentration, then they produce different amount of H⁺. Acids that give rise to more H⁺ ions are said to be strong acids and acids that give less H⁺ ions are said to be weak acids.

6. Importance of pH in everyday life:

(i) If pH of rain water is less than 5.6, it is called acid rain. When acid rain flows into the rivers, it lowers the pH of river water. The survival of aquatic life in such rivers become difficult. Acid rain also damage crops and cause a change in pH of the soil.

(ii) pH in our digestive system: Our stomach produces digestive juices/hydrochloric acid (HCl), which helps in the digestion of food without harming the stomach. However, sometimes the stomach produces too much of acid and this causes indigestion, which is accompanied by pain and irritation. To get rid of this pain, people use antacids like magnesium hydroxide. These antacids neutralise the excess acid formed.

(iii) pH change as the cause of tooth decay: Tooth decay starts when the pH of the mouth is lower than 5.5. Tooth enamel, made up of calcium phosphate is the hardest substance in the body. It does not dissolve in water but is corroded when the pH in the mouth is below 5.5. Using toothpaste, which are generally basic, for cleaning the teeth can neutralise the excess acid and prevent tooth decay.

(iv) Bee-sting leaves an acid which causes pain and irritation. Using a mild base like baking soda on the stung area gives relief. Stinging hair of nettle leaves inject methanoic acid causing a burning pain. A traditional remedy is rubbing the area with the leaf of the dock plant.

(v) Various fluids in our body work within a particular range of pH such as, pH of human blood should be between 7.3 to 7.5.

(vi) For the growth of plants, a particular pH range of soil is essential. Usually neutral soil is best for crops. If the soil is acidic, farmers treat the soil with quick lime or slaked lime.

(vii) The tarnished surface of a copper vessel due to the formation of copper oxide layer (which is basic) can be cleaned by rubbing with lemon (which is acidic).

7. Salts: Salts are generally ionic compounds which are obtained by neutralisation reaction between acids and bases.

Acid + Base→ Salt + Water
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(Z)

In these salts, the cation is derived from base and anion is derived from acid.
Salts are mostly solids with high melting points. They are soluble in water.

Different types of salts:

• Normal salts : NaNO₃
• Acidic salts : NaHSO₄
• Basic salts : Pb(OH)Cl
• Double salts : FeSO₄.(NH₄)₂.SO₄.6H₂O

8. Remember: Salts of a strong acid and a strong base are neutral with pH value of 7. On the other hand, salts of strong acid and weak base are acidic with pH values less than 7 and those of strong base and weak acid are basic in nature, with pH value more than 7.

Common Salt (NaCl) [Table salt]
Sodium chloride (NaCl) also called common salt or table salt is the most common/essential part of our diet. It is obtained by neutralisation reaction of sodium hydroxide (NaOH) with hydrochloric acid (HC1).

It is obtained on a large scale from sea water.
Large crystals are often brown due to the presence of impurities. This is called rock salt. Beds of rock salt were formed when seas of by gone ages dried up. Rock salt is mined like coal.

9. Uses of Sodium Chloride

• Sodium chloride is a major ingredient of edible salt.
• It is used as a food preservative.
• Compounds like sodium hydroxide (NaOH), baking soda (NaHCO₃) and washing soda are obtained from sodium chloride.
• It is used to melt ice on hill stations and cold countries during heavy snow fall.

10. Sodium Hydroxide (NaOH): Sodium hydroxide (NaOH) is commonly known as caustic soda.

Sodium hydroxide is manufactured by electrolysis of an aqueous solution of sodium chloride (called brine). Chlorine gas is given off at the anode and hydrogen gas at the cathode. Sodium hydroxide solution is formed near the cathode.
2NaCl(aq) + 2H₂O(l) → 2NaOH(ag) + Cl₂(g) + H₂(g)

The process is called the chlor-alkali process because of the products formed chlor for chlorine and alkali for sodium hydroxide.

11. Uses of Sodium Hydroxide

1. Sodium hydroxide is used for making soaps and detergents.
2. Sodium hydroxide is used for making artificial textile fibres (such as rayon).
3. It is used the preparation of soda lime (a mixture of NaOH and CaO).
4. It is used as a cleansing agent for machines and metal sheets.

12. Baking Soda (NaHCO_s): The chemical name of baking soda is sodium hydrogencarbonate or sodium bicarbonate (NaHCO₃). It can be prepared from sodium chloride as
NaCl + H₂O + CO₂ + NH₃→ NH₄Cl + NaHCO₃

Since it is slightly soluble in water, it can be removed by filtration.
It is a mild non-corrosive base. The following reaction takes place when it is heated during cooking.

13. Uses of Baking Soda
Being alkaline it is an ingredient in antacids. It neutralises excess acid in the stomach and provides relief.
NaHCO₃ + HCl → NaCl + H₂O + CO₂

• It is used in soda-acid fire extinguisher.
• It is used in making baking powder (a mixture of baking soda and mild edible acid like tartaric acid). When baking powder is heated or mixed in water CO₂ gas is released.
  NaHCO₃ + H⁺→ CO₂ + H₂O + Sodium salt of acid

The released CO₂ causes breads or cakes to rise making them soft and spongy/fluffy.

14. Bleaching Powder [CaOCl₂]: Bleaching powder is calcium oxychloride. It is also known as chloride of lime. Bleaching powder can be prepared by the action of chlorine on dry slaked lime [Ca(OH₂)].
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O
Bleaching powder is a yellowish white solid.

15. Uses of Bleaching Powder

• It is an oxidising agent.
• It is used for disinfecting drinking water to make it free from germs.
• The most important use of bleaching powder is in:
• textile industry for bleaching cotton and linen
• paper industry for bleaching wood pulp
• laundry for bleaching washed clothes.

16. Washing Soda [Na₂Og.10H₂O]: Sodium carbonate is obtained by heating baking soda. When the sodium carbonate obtained by the above process is recrystallised, we get washing soda.
Na₂CO₃ + 10H₂O → Na₂CO₃.10H₂O
Anhydrous sodium carbonate is called soda ash.

17. Uses of Washing Soda

• It is used in glass, soap and paper industries.
• It is used in the manufacture of borax.
• It is used as a cleaning agent for domestic purposes.
• It is used for removing permanent hardness of water.

18. Plaster of Paris \(\left(\mathrm{CaSO}_{4} \cdot \frac{1}{2} \mathrm{H}_{2} \mathrm{O}\right)\)
Plaster of Paris is calcium sulphate hemihydrate, it can be obtained by heating gypsum at 373 K.
Plaster of Paris is a white powder and on mixing with water, it changes to gypsum once again giving a hard solid mass.

19. Knowledge Booster
In washing soda, (Na₂CO₃.10H₂O), 10H₂O signify water of crystallisation. Water of crystallisation is the fixed number of water molecules present in one formula unit of a salt. Some other examples are

20. Uses of Plaster of Paris

• Plaster of Paris is used by doctors as plaster for supporting fractured bones in the right position.
• It is also used for making toys, materials for decoration and for making surfaces smooth.

Class 10 Science Chapter 2 Notes Important Terms

Alkalies are water soluble bases.

Rock salt is chemically sodium chloride (NaCl).

Antacid is a substance which can neutralise acidity in the stomach.

Neutralisation is the reaction in which an acid reacts with a base to form salt and water.

Bleaching powder is chemically calcium oxychloride (CaOCl₂) and is formed by passing chloride gas through dry slaked lime.

Amphoteric compound is a compound that can act both as an acid and a base.

Dilute Acid: Contains only a small amounts of acid and a large amount of water.

Concentrated Acid: A concentrated acid contains a large amount of acid and a small amount of water.

On this page, you will find Areas related to Circles Class 10 Notes Maths Chapter 12 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 12 Areas related to Circles will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Maths Chapter 12 Notes Areas related to Circles

Areas related to Circles Class 10 Notes Understanding the Lesson

1. Circle: Circle is set of all points in a plane which are at the fixed distance from a fixed point i.e., centre. Centre: Mid-point of a circle is called centre of a circle.

2. Radius: The distance between the centre of a circle to the circumference of the circle.
It is denoted by r or R.

3. Chord: A line segment which joins the two points in the circumference of the circle.

4. Diameter: It is the longest chord which passes through the centre of a circle. It is denoted by d or D.
Diameter = 2 x radius
\(\text { Radius }=\frac{\text { Diameter }}{2}\)

5. Circumference of a Circle (or Perimeter)
A perimeter is a path that surrounds a two dimensional shape.
The perimeter of a circle is called its circumference. Circumference
\(\frac{\text { Circumference }}{\text { Diameter }}=\pi\)
Circumference = n x diameter = π x 2r = 2πr
or
The value of π is \(\frac{22}{7}\) or 3.14 (approximately)

6. Arc: An arc is the part of the circumference of a circle.
An arc AB is denoted as \(\widehat{\mathrm{AB}}\)

Length of arc AB is l \(\widehat{\mathrm{AB}}\) or l.
Length of an arc of a sector of angle
\(\theta=\frac{2 \pi r \theta}{360^{\circ}}\)

7. Sector: The portion (or part) of circular region enclosed by two radii and the corresponding arc is called a sector of the circle.

8. Minor and Major sector of the circle: Shaded region OAPB is called sector or minor sector of a circle with centre O.
∠AOB is called angle of the sector.

And OAQB is called major sector.
Angle of major sector = 360° – ∠AOB.

9. Area of Circle

Let r be the radius of circle. If we cut the circle in sectors and arrange then we see this figure like a rectangle whose length is \(\frac{1}{2}\) 2πr = πr and breadth is r.
Hence,
Area of circle = Area of rectangle
= l x b = \(\frac{1}{2}\) 2πr = r
Area of circle = πr²

10. Segment: The portion or part of a circular region enclosed between a chord and the corresponding arc is called a segment of the circle.
The shaded region APB is the minor segment.
And the region AQB is the major segment.
Area of circle = πr²
Area of the sector of an angle \(\theta=\frac{\pi r^{2} \theta}{360^{\circ}}\)

Area of major sector = Area of circle – Area of minor sector
Area of minor segment = Area of sector – Area of ΔOAB\(\frac{\pi r^{2} \theta}{360^{\circ}}\)– area of ΔOAB
Area of major segment = Area of circle – Area of minor segment

11. Area of Combinations of Plane Figures
In our daily life we have observed various plane figures which are combinations of two or more figures and i also in the form of various interesting designs like flower beds, curtains, drain covers, window designs, ] designs on table covers.  To calculate areas of such figures, we field the area of shapes used and then and/subtract as per need.

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CBSE Class 10 Science Chapter 1 Notes Chemical Reactions and Equations

Chemical Reactions and Equations Class 10 Notes Understanding the Lesson

1. Chemical reaction: The reaction in which the original state of the particles changes and it cannot be reversed by simple physical means, is known as a chemical reaction.
Examples: fermentation of grapes, burning of wood, etc. Burning of wood produces charcoal and we cannot get back wood from charcoal on reversing the conditions.

• Chemical reaction is accompanied by change in state, colour, evolution of gas or change in temperature. The chemical reaction is represented as
  Reactants → Products
• Example of a chemical reaction is burning of magnesium ribbon with a dazzling white flame to form a white powder (magnesium oxide).
  2Mg + O₂ → 2MgO

2. Chemical equation: Representation of a chemical reaction in terms of chemical symbols and formulae of the reactants and products is known as chemical equation. A chemical equation represents the reactants, products and their physical states symbolically.
For example,
Magnesium + Oxygen → Magnesium oxide
2Mg(s) + O₂(g) → 2MgO(s)
The substances that undergo chemical change in the reaction, i.e., magnesium and oxygen, are the reactants. The new substance, magnesium oxide, formed during the reaction is the product.

Writing a chemical reaction in terms of chemical equation: A chemical equation represents a chemical reaction. While writing a chemical equation the reactants are written on the left hand side of the equation while products on the right hand side.
Examples:

3. Balanced chemical equation: The chemical equation in which the number of atoms of different elements is same on both sides of the arrow is called a balanced chemical equation.
This is in accordance to the law of conservation of mass.
Let us try to balance the following chemical equation:
Fe + H₂O → Fe₃O₄ + H₂
Number of atoms of different elements present in the unbalanced equation

Element	Number of atoms in reactants (LHS)	Number of atoms in products (RHS)
Fe	1	3
H	2	2
O	1	4

Now select the element which has the maximum number of atoms Fe304. There are four oxygen atoms on the RHS and only one on the LHS.

To balance the oxygen atoms

Atoms of oxygen	In reactants	In products
(i) Initial	1 (in H2O)	4 (in Fe3O4)
(ii) To balance	1 x 4	4

So, multiplying H₂O molecules by four, we get
Fe + 4H₂O → Fe₃O₄ + H₂

To balance H atoms, make the number of molecules of hydrogen as four on the RHS.
Fe + 4H₂O → Fe₃O₄ + 4H₂

To equalise Fe, we multiply Fe atoms by three on the LHS.
Hence,
3Fe + 4H₂O → Fe₃O₄ + 4H₂ (Balanced equation)

To make chemical equation more informative gaseous, liquid, aqueous and solid states of reactants and products are represented by the notations (g), (l), (aq) and (s) respectively.
Hence,  3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)

4.Types of chemical reactions
(a) Combination reaction: The reactions in which two or more substances combine to form a new substance is called combination reaction.
For example,

• 2Mg(s) + O₂ (fe)→ 2MgO(s)
• CaO(s) + H₂O(l) → Ca(OH)₂(aq)

(b) Decomposition reaction: The reaction in which a single compound breaks up into two or more simpler substances is called decomposition reaction. For example,

2Pb(NO₃)₂(s) → 2PbO(s) + 4NO₂(g)+ O₂(g)

The decomposition of a substance by passing electric current through it is known as electrolysis.

The decomposition of a substance on heating is known as thermal decomposition.

The decomposition of a substance by absorbing light energy is called photochemical decomposition.
For example,

The above two reactions are used in black and white photography.

Decomposition reactions are opposite of combination reactions.

(c) Displacement reaction: The chemical reaction in which a more reactive element displaces a less reactive element from its salt solution is known as displacement reaction. For example,

• Zn(s) + CuSO₄(aq)→  ZnSO₄(aq) + Cu(s)
• Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂ (aq) + 2Ag(s)

(d) Double displacement reaction: In this reaction two different atoms or group of atoms are mutually exchanged.

A white, insoluble substance, i.e., BaS04 is formed which is called precipitate.
Precipitation Reaction-Any reaction that produces a precipitate is called a precipitation reaction.

(e) Oxidation: Oxidation is the gain of oxygen or loss of hydrogen.

(f) Reduction: Reduction is the loss of oxygen or gain of hydrogen.

Redox reaction: The reaction in which one reactant gets oxidised while other gets reduced.

Exothermic reactions: Reaction in which heat is released along with the formation of products.
For example, \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(g)\)
Respiration and decomposition of vegetable matter into compost are exothermic reactions.

Endothermic reactions: The reactions which require energy in form of heat, light or electricity are called endothermic reactions.
\(2 \mathrm{Ba}(\mathrm{OH})_{2}+2 \mathrm{NH}_{4} \mathrm{Cl} \longrightarrow \text { Heat } \mathrm{2BaCl}_{2}+2 \mathrm{NH}_{4} \mathrm{OH}\)

5. Corrosion: The process of slow deterioration of some metals like iron, copper and silver into their compounds due to their reaction with oxygen, water, acids, gases, etc. present in the atmosphere is called corrosion.
Rusting The process in which iron reacts with oxygen and moisture present in the air to form a reddish brown coating called rust on its surface.

6. Rancidity: The taste and odour of food materials containing fat and oil changes when they are left exposed to air for a long time. This is called rancidity. It is caused due to oxidation of fat and oil present in food material. It can be prevented by using various methods such as by adding antioxidants to the food materials, storing food in air tight containers and by flushing out air with nitrogen.

Class 10 Science Chapter 1 Notes Important Terms

Chemical reaction is a process in which old bond breaks up and new bonds are formed.

Chemical equation is the representation of a chemical reaction in terms of chemical symbols and formulae.

Combination reaction is a reaction in which two or more substances combine to form a new substance.

Decomposition reaction is a reaction in which a single compound breaks up into two or more simpler substances.

Displacement reaction is a reaction in which a more reactive element displaces a less reactive element from its salt solution.

Redox reaction is the reaction in which oxidation and reduction takes place simultaneously.

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CBSE Class 10 Maths Chapter 11 Notes Constructions

Constructions Class 10 Notes Understanding the Lesson

Division of a line segment internally in the given ratio.

Let AB be a line segment of certain length. We need a point P on AB dividing it internally in the ratio m : n
Steps of Construction: Let m = 4, n = 3.

• Draw a line segment AB of given length.
• Make an acute ∠BAX with AB
• Use a compass of any radius and mark 7(i.e. m + n) points A₁, A₂,…… , A₇
  such that AA₁ = A₁A₂ = A₂A₃ =………………= A₆A₇
  
  
• Join BA₇
• Through the point A₄ [m = 4], draw a line parallel to BA₇ by making an angle equal to AA₇B at A₄ intersecting AB at P. Then AP : PB = 4 : 3.

Construction of a Triangle similar to a given Triangle as per given scale factor.
1. Scale Factor  \(\frac{m}{n}\)(where m < n)
Steps of construction: Let \(\frac{m}{n}=\frac{3}{4}\)

• Construct a triangle ABC by using given data.
• Make an acute angle ∠BAX below the base AB.
• Along AX, mark 4 points [the greater of 3 and 4 in \(\frac{3}{4}\)] as A₁, A₂,A₃, A₄ such that
  AA₁ = A₁A₂ = A₂A₃ = A₃A₄.
• Join A₄ B
  
• From A₃, draw A_gB’ || A₄B, meeting AB at B’.
  From B’, draw B’C’ || BC, meeting AC at C’.

Thus, ΔAB’C’ is the required triangle each of whose sides is \(\frac{3}{4} \)of the corresponding side of ΔABC.

2. Scale Factor \(\frac{m}{n} \) (where m > n)
Steps of construction:
Let \(\frac{m}{n}=\frac{5}{3}\)

• Construct ΔABC using given data.
• Make an acute angle ∠BAX below the base AB. Extend AB to AY and AC to AZ.

• Along AX, mark 5 points [the greater of 5 and 3 in \(\frac{5}{3} \)
  such that AA₁ = A₁A₂ =………. = A₄A₅.
• Join A₃B.
• From A_g, draw A₅B’ || A₃B, meeting AY produced at B’.
• From B’, draw B’C’ || BC, meeting AZ produced at C’.
  Thus, ΔAB’C’ is the required triangle, each of whose sides is \(\frac{5}{3}\) of the corresponding side of ΔABC.

Construction of the pair of tangents from an external point to a circle.

Let O is the centre of the circle and a point A is external point to a circle.

Steps of construction

• Join AO and bisect it. Let M be the mid-point of AO.
• Taking M as centre and MO as radius, draw a circle.
  Let it intersects the given circle at the points B and C.
• Join AB and AC.
  Thus, AB and AC are the required tangents.

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CBSE Class 10 Maths Chapter 10 Notes Circles

Circles Class 10 Notes Understanding the Lesson

1. Circle: A circle is a collection of all points in a plane which are at a constant distance from a fixed point. Here
Fixed point is called centre of the circle. Constant distance is called radius of the circle.

2. Considering a circle with centre ‘O’ and radius V and a line T in a plane.
Three different situations are there:

• When there is no common point between the circle and the line. Then the line is known as a non¬intersecting line
• When the line passes the circle in two points, the line is called a secant
• When a line meets the circle at a point, the line is called a tangent
  The point at which the tangent touches the circle is called point of contact.

Facts related to tangent of circle.

Given: A circle with centre 0 and radius r.
A tangent XY at point P to the circle.
To prove: OP ⊥ XY
Construction: Take a point Q on XY other than P. Join OQ.
Proof: Point Q lies outside the circle.

If point Q lies inside the circle then XY will become a secant and not a tangent to the circle.
∴ OQ > OP
which is true for every point on the line XY except the point P.
⇒ OP is the shortest of all the distances of the point O to the points of XY.
OP ⊥ XY
[ ∵ Shortest length from the point outside the line to the line is perpendicular]
Remark: A line drawn through the end point of radius and perpendicular to it, is the tangent to the circle.

Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.
Given: PT and PS are tangents from point P to circle with centre 0.
To prove: PT = PS
Construction: Join OP, OT and OS.
Proof: In ΔOTP and ΔOSP
OT = OS [Radii of same circle]

OP = QP
∠OTP = ∠OSP [Each 90º]
∴ ΔOTP ≅ ΔOSP  [RHS]
⇒ PT = PS   [CPCT]

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CBSE Class 10 Maths Chapter 9 Notes Some Applications of Trigonometry

Some Applications of Trigonometry Class 10 Notes Understanding the Lesson

Trigonometry is the study of relationships between the sides and angles of a triangle. In this chapter you will study about some ways in which trigonometry is used:

• It is used in geography and in navigation.
• It is used in constructing maps, determine the position of an island in relation to the longitudes and latitudes.
• It is used for calculating the height and distance of various objects without measuring it.

Terms related to height and distance:

1. Line of sight: The line joining the eyes of the observer and the objects which he/she observes is called line of sight.

2. Angle of elevation: (When object is above the horizontal)
The angle between the line of sight and the horizontal is called the angle of elevation.

3. Angle of depression: (When object is below the horizontal)
The angle between the horizontal line and the line of sight is called the angle of depression.

Trigonometric formulae used:

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CBSE Class 10 Maths Chapter 6 Notes Triangles

Triangles Class 10 Notes Understanding the Lesson

In X standard we have learnt about congruent figures.

Congruent figure: Those two geometric figures having the same shape and size are known as congruent figures.

Rules of Congruency

1. SAS (Side-Angle-Side): Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

2. ASA (Angle-Side-Angle): Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.

3. AAS (Angle-Angle-Side): Two triangles are Congruent if any two pairs of angles and a pair of corresponding sides are equal.

4. SSS (Side-Side-Side): If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

5. RHS (Right angle-Hypotenuse-Side: In two right angle triangles, the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

Note: All congruent figures or triangles are similar.

Similar Figure: Two figure which are of same shape (but not necessarily the same size) are called similar figures. For example,

• All line segments are similar.
• All circles are similar.
• Two or more squares are similar.
• Two or more equilateral triangles are similar.

Note:

• All rectangles are not similar.
• All triangles are not similar.

Similar Polygons: Two polygons with the same number of sides are similar, if (1) their corresponding angles are equal. (2) their corresponding sides in the same ratio.

Similarity of Triangles

Two triangles are similar if

• Their corresponding angles are equal; and
• Their corresponding sides are in the same ratio

Famous Greek mathematician Militus Thales gives the relation to the two equiangular triangle is known as BPT or Thales theorem.

Equiangular triangles: If corresponding angles of two triangles are equal then they are equiangular triangles

Theorem 6.1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given: A AABC in which a line DE || BC intersects the other two sides AB and AC at D and E respectively.
To Prove that \(\frac{A D}{D B}=\frac{A E}{E C}\)
Construction: Join BE and CD and draw DM ⊥ AC and EN ⊥ AB

(Because both are on the same base DE and between the same parallels BC and DE) from eqn (1), (2) and (3) AD AE

Theorem 6.2: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

AD AE Given: In ΔABC, \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
To prove: DE || BC

Construction: Let us suppose that DE is not parallel to BC, so we draw a line DF || BC
Proof: DF || BC
Therefore by Basic Proportionality Theorem,

which is not possible. We come at the contradiction. So our supposition was wrong, it is only possible, if point F will coincide the point E.
Therefore DE || BC.

Criteria for Similarity of Triangles

In previous section, we have studied that two triangles are similar, if (I) their corresponding angles are similar (II) their corresponding sides are proportional (or are in the same ratio).

Theorem 6.3: AAA Criterion: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

In ΔABC and ΔDEF,

Remark : AA Similarity Creterian: If two angles of a triangle are equal to two angles of another triangle, then their corresponding angles are equal and the triangles are similar.

In ΔABC and ΔDEF
∠A =∠D, ∠C = ∠F then ΔABC ~ ΔDEF (by AA similarity)

Theorem 6.4: SSS Similarity Criterion: If the corresponding sides of two triangles are proportional (i.e. in the same ratio), then their corresponding angles are equal and the triangles are similar.

In ΔABC and ΔDEF
\(\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F} DF \)then ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
Hence ΔABC ~ ΔDEF

Theorem 6.5: SAS Similarity Criterion: If one angle of a triangle is equal to one angle of the other and the sides including these angles are proportional, the triangles are similar.

In ΔABC and ΔDEF
∠BAC = ∠EDF
\(\frac{A B}{D E}=\frac{A C}{D F}\)
Hence ΔABC ~ ΔDEF

Areas of Similar Triangles
We have study in two similar triangles. Ratio of the corresponding sides of two similar triangles is same.

Theorem 6.6: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: ΔABC ~ ΔPQR

Theorem 6.7: If a perpendicular is drawn from the verities of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

• ΔADB ~ ΔABC
• ΔBDC ~ ΔABC
• ΔADB ~ ΔBDC

Theorem 6.8: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given: ABC is a right angle triangle which is right angled at B.
To prove:  AC² = AB² + BC²
Construction: Draw BD ⊥  AC
Proof:  ΔADB ~ ΔABC
(If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other)

From equation (2) and (3)
Adding eqn (1) and (2)
(If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse than triangles on both sides of the perpendicular are similar to the whole triangle and to each other)
AB² + BC² = AD . AC + CD . AC
⇒ AB² + BC² = AC (AD + CD)
⇒ AB² + BC² = AC x AC
⇒ AB² + BC² = AC²
⇒ AC² = AB² + BC²

Theorem 6.9: In a triangle, if square of one side is equal to the sum of the squares of the other two sides. Then the angle opposite the first side is a right angle.

Given: We have ΔABC in which
AC² = AB² + BC²
To prove: ∠ABC = 90°
Construction: Construct a ΔDEF right angled at E such that EF = BC and DE = AB

Proof: In ΔDEF
DF² = EF² + DE² (Pythagoras theorem) (given)
DF² = BC² + AB²…(1) (by construction)
But AC² = BC² + AB²…(2)

From eqn (1) and (2)
AC² = DF²
⇒ AC = DF… (3)

In ΔABC and ΔDEF
AB = DE(by construction)
BC = EF(by construction)
AC = DF(proved above in eqn (3))
ΔABC ≅ ΔDEF(by SSS congruence)
⇒ ∠ABC = ∠DEF (by CPCT)
Therefore ∠ABC = 90°

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CBSE Class 10 Maths Chapter 8 Notes Introduction to Trigonometry

Introduction to Trigonometry Class 10 Notes Understanding the Lesson

The word trigonometry is derived from the Greek words ‘Tri’ which means three, ‘gon’ means sides and metron meaning measure.

It means trigonometry is the study of relationship between the sides and angles

• The earliest work on trigonometry was recorded in Egypt and Babylon.
• Trigonometry was used by early astronomers to find out the distance of stars and planets from the earth.

Trigonometric Ratios

The ratios of the sides of a right triangle with respect to its acute angles are called trigonometric ratios.

1. In right triangle, side opposite to given acute angle will always be perpendicular.
2. Side opposite 90° will always be hypotenuse.
3. Remaining side will be base.
4. The sum of two angles (except right angle) is 90°
   i.e.,      ∠A + ∠C = 90°           ( ∵ ∠B = 90°)
   

1. sin θ = sin θ = \(\frac{0}{\mathrm{H}}\) (O- side opposite to given angle i.e., acute angle)

2. cosine θ = cos θ = \(\frac{\mathrm{A}}{\mathrm{H}} \)(A-adjacent side) (H-Hypotenuse)

3. Tangent θ = tan θ =\(\frac{\mathrm{O}}{\mathrm{A}}\)
(O-side opposite to acute angle, A-adjacent side)

4. cosecant θ = cosec θ= \(\frac{\mathrm{H}}{\mathrm{O}}\)

5. secant θ = sec θ =\(\frac{\mathrm{H}}{\mathrm{A}}\)

6. cotangent θ= cot θ =\(\frac{\mathrm{A}}{\mathrm{O}}\)

Trigonometric angles for some specific angles

Also we can find values for some special angles as follows:

Trigonometric Identities

Identity: That equation is called an identity. If it is true for all values of the variables which involved. I. In right ΔABC,