Find the values of: sin ∠ABC

(a) From the figure (1) given below, find the values of:

(i) sin ∠ABC

(ii) tan x – cos x + 3 sin x.

(b) From the figure (2) given below, find the values of:

(i) 5 sin x

(ii) 7 tan x

(iii) 5 cos x – 17 sin y – tan x.

Answer :

(a) BC = 12, CD = 9 and BC = 20

In right angled ∆ABC,

AB² = AC² + BC²

AC² = AB² – BC²

AC² = (20)² – (12)²

AC² = 400 – 144 = 256

So we get

AC² = (16)²

⇒ AC = 16

In right angled ∆BCD

BD² = BC² + CD²

BD² = 12² + 9²

BD² = 144 + 81 = 225

BD² = (15)²

⇒ BD = 15

(i) In right angled ∆BCD

sin ∠ABC = perpendicular/hypotenuse

sin ∠ABC = AC/AB = 16/20 = 4/5

(ii) In right angled ∆BCD

tan x = perpendicular/base

tan x = BC/CD = 12/9 = 4/3

In right angled ∆BCD

cos x = base/hypotenuse

cos x = CD/BD = 9/15 = 3/5

In right angled ∆BCD

sin x = perpendicular/hypotenuse

sin x = BC/BD = 12/15 = 4/5

⇒ tan x – cos x + 3 sin x = 4/3 – 3/5 + (3× 4/5)

= 4/3 – 3/5 + 12/5

Taking LCM

= [(4×5) – (3×3) + (12×3)]/15

= (20 – 9 + 36)/15

= (56 – 9)/15

= 27/15

= 3 (2/15)

Hence, tan x – cos x + 3 sin x = 3 2/15.

(b) AC = 17, AB = 25, AD = 15

In right angled ∆ACD

AC² = AD² + CD²

(17)² = (15)² + (CD)²

CD² = (17)² – (15)²

⇒ CD² = 289 – 225 = 64

CD² = 8²

⇒ CD = 8

In right angled ∆ABD

AB² = AD² + BD²

(25)² = (15)² + BD²

BD² = (25)² – (15)²

⇒ BD² = 625 – 225 = 400

So we get

BD² = (20)²

⇒ BD = 20

(i) In right angled ∆ABD

5 sin x = 5 (perpendicular/hypotenuse)

= 5 (AD/AB)

= 5 × 15/25

= 15/5

= 3

(ii) In right angled ∆ABD

7 tan x = 7 (perpendicular/base)

= 7 (AD/AB)

= 7× 15/20

= 7× ¾

= 21/4

= 5 ¼

(iii) In right angled ∆ABD

cos x = base/hypotenuse

cos x = BD/AB = 20/25 = 4/5

In right angled ∆ACD

sin y = perpendicular/hypotenuse

sin y = CD/AC = 8/17

In right angled ∆ABD

tan x = perpendicular/base

tan x = AD/BD = 15/20 = ¾

5 cosx – 17 siny – tanx = (5× 4/5) – (17× 8/17) – ¾

It can be written as

= 4/1 – 8/1 – ¾

Taking LCM

= (16 – 32 – 3)/4

= (16 – 35)/4

= –19/4

= –4 ¾

Hence, 5 cos x – 17 sin y – tan x = – 4 ¾.

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