If AB = y units, BC = 3 units and CA = 5 units, find

(a) From the figure (1) given below, find the values of:
(i) 2 sin y- cos y
(ii)2 sinx-cos x
(iii) 1- sin x + cos y
(iv) 2 cos x-3 sin y +4 tan z.

(b) In the figure (2) given below, ∆ABC is right-angled at B. If AB = y units, BC = 3 units and CA = 5 units, find

(i) sin x°

(ii) y.

Answer :

(a) In a right angled ∆BCD,

BC² = BD² + CD²

BC² = 9² + 12²

BC² = 81 + 144 = 225

⇒ BC² = 15²

⇒ BC = 15

In a right angled ∆ABC,

AC² = AB² + BC²

AB² = AC² – BC²

AB² = 25² – 15²

AB² = 625 – 225 = 400

So we get

AB² = 20²

AB = 20

(i)

In right angled ∆BCD

sin y = perpendicular/ hypotenuse

⇒ sin y = BD/ BC

sin y = 9/15 = 3/5

In right angled ∆BCD

cos y = base/hypotenuse

⇒ cos y = CD/BC

cos y = 12/15 = 4/5

2sin y – cos y = 2× 3/5 – 4/5

= 6/5 – 4/5

= 2/5

Therefore, 2 sin y – cos y = 2/5

(ii) In right angled ∆ABC

sin x = perpendicular/ hypotenuse

sin x = BC/AC

sin x = 15/25 = 3/5

In right angled ∆ABC

cos x = base/hypotenuse

⇒ cos x = AB/AC

cos x = 20/25 = 4/5

2 sin x – cos x = 2× 3/5 – 4/5

= 6/5 – 4/5

= 2/5

Therefore, 2 sin x – cos x = 2/5.

(iii) In right angled ∆ABC

sin x = perpendicular/hypotenuse

⇒ sin x = BC/AC

sin x = 12/25 = 3/5

In right angled ∆BCD

cos y = base/hypotenuse

⇒ cos y = CD/BC

cos y = 12/15 = 4/5

1 – sin x + cos y

= 1 – 3/5 + 4/5

= (5 – 3 + 4)/ 5

So we get

= (9 – 3)/ 5

= 6/5

Therefore, 1 – sin x + cos y = 6/5.

(iv) In right angled ∆BCD

cos x = base/hypotenuse

⇒ cos x = AB/AC

cos x = 20/25 = 4/5

In right angled ∆BCD

sin y = perpendicular/hypotenuse

⇒ sin y = BD/BC

sin y = 9/15 = 3/5

In right angled ∆ABC

tan x = perpendicular/base

⇒ tan x = BC/AB

tan x = 15/20 = ¾

Here,

2 cos x – 3 sin y + 4 tan x = (2× 4/5) – (3× 3/5) + (4× ¾)

= 8/5 – 9/5 3/1

Taking LCM

= (8 – 9 + 15)/5

= (23 – 9)/ 5

= 14/5

(b)  AB = y units, BC = 3 units, CA = 5 units

(i) In right angled ∆ABC

sin x = perpendicular/hypotenuse

⇒ sin x = BC/AC

sin x = 3/5

(ii) In right angled ∆ABC

AC² = BC² + AB²

We can write it as

AB² = AC² – BC²

AB² = 5² – 3²

AB² = 25 – 9 = 16

So we get

AB² = 4²

⇒ AB = 4

y = 4 units

Hence,

y = 4 units.

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