## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test

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- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19
- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test

**Question 1.**

**Using trigonometrical tables, find the values of :**

**(i) sin 48° 52′**

**(ii) cos 37° 34′**

**(iii) tan 18° 21′.**

**Solution:**

Using tables, we find that

(i) sin 48° 52′ = .7524 + .0008 = .7532

(ii) cos 37° 34′ = .7934 – .0007 = .7927

(iii) tan 18° 21′ = .3307 + .0010 = .3317.

**Question 2.**

**Use tables to find the acute angle θ, given that**

**(i) sin θ = 0.5766**

**(ii) cos θ = 0.2495**

**(iii) tan θ = 2.4523.**

**Solution:**

Using table, we find that

(i) sin θ = 0.5766 = 0.5764 + 0.0002

= sin (35° 12’+ 1′)

= sin 35° 13′

θ = 35° 13′

(ii) cos θ = 0.2495 = 0.2487 + 0.0008

= cos (75° 36′ – 3′)

= cos 75° 33′

θ = 75° 33′

(iii) tan θ = 2.4523 = 2.4504 + 0.0019

= tan (67° 48′ + 1′)

= tan 67° 49′

**Question 3.**

**If θ is acute and cos θ = 0.53, find the value of tan θ.**

**Solution:**

From the table, we find that

cos θ = 0.53 = .5299 + .0001 = cos 58°

θ = 58°

and tan 58° = 1.6003

**Question 4.**

**Find the value of: sin 22° 11′ + cos 57° 20′ – 2 tan 9° 9′.**

**Solution:**

Using the tables, we find that

sin 22° 11′ = 0.3762 + 0.0014 = 0.3776

cos 57° 20′ = 0.5402 – 0.0005 = 0.5397

tan 9° 9′ = 0.1602 + 0.0009 = 0.1611

∴ sin 22° 11′ + cos 57° 20′ – 2 tan 9° 9′

= 0.3376 + 0.5397 – 0.1611 × 2

= 0.3776 + 0.5397 – 0.3222

= 0.9173 – 0.3222

= .5951.

**Question 5.**

**If θ is acute and sin θ = 0.7547, find the value of: (i) θ (ii) cos θ (iii) 2 cos θ – 3 tan θ.**

**Solution:**

Using the tables, we find that

(i) sin θ = 0.7547 = sin 49°

θ = 49°.

(ii) cos θ = cos 49° = 0.6561.

(iii) tan θ = tan 49° = 1.1504

2 cos θ – 3 tan θ

= 2 × θ.6561 – 3 × 1.1504

= 1.3122 – 3.4512

= – 2.1390

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