**(a) In figure (1) given below, ∆ABC is right-angled at B and ∆BRS is right-angled at R. If AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x° and ∠SAB = y°, then find:**

(i) tan x°

(ii) sin y°.

**(b) In the figure (2) given below, ∆ABC is right angled at B and BD is perpendicular to AC. Find**

(i) cos ∠CBD

(ii) cot ∠ABD.

**Answer :**

**(a)** ∆ABC is right-angled at B, ∆BSC is right-angled at S and ∆BRS is right-angled at R

AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x° and ∠SAB = y°

By Geometry ∆ARS and ∆ABC are similar

AR/AB = RS/BC

AR/18 = 5/7.5

AR = (5×18)/7.5 = (1×18)/1.5

Multiply both numerator and denominator by 10

AR = (18×10)/15

⇒ AR = (10×6)/5

⇒ AR = (2×6)/1 = 12

RB = AB – AR

⇒ RB = 18 – 12 = 6

In right angled ∆ABC

AC^{2} = AB^{2} + BC^{2}

AC^{2} = 18^{2} + 7.5^{2}

AC^{2} = 324 + 56.25 = 380.25

⇒ AC = √380.25 = 19.5 cm

**(i)** In right angled ∆BSR

tan x° = perpendicular/base

tan x° = RB/RS = 6/5

**(ii)** In right angled ∆ASR

sin y° = perpendicular/hypotenuse

AS^{2} = 12^{2} + 5^{2}

AS^{2} = 144 + 25 = 169

⇒ AS = √169 = 13 cm

sin y° = RS/AS = 5/13

**(b) **∆ABC is right angled at B and BD is perpendicular to AC

In right angled ∆ABC

AC^{2} = AB^{2 }+ BC^{2}

AC^{2} = 12^{2} + 5^{2}

AC^{2} = 144 + 25 = 169

AC^{2} = (13)^{2}

⇒ AC = 13

By Geometry ∠CBD = ∠A and ∠ABD = ∠C

**(i) **cos ∠CBD = cos ∠A = base/hypotenuse

In right angled ∆ABC

cos ∠CBD = cos ∠A = AB/AC = 12/13

**(ii)** cos ∠ABD = cos ∠C = base/perpendicular

In right angled ∆ABC

cos ∠ABD = cos ∠C = BC/AB = 5/

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