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°.

In ∆ARS and ∆ABC,

∠ARS = ∠ABC (Both = 90°)

∠SAR = ∠CAB (Common)

∴ ∆ARS ~ ∆ABC [By AA axiom]

∴ $\dfrac{AR}{AB} = \dfrac{RS}{BC}$

Substituting the values we get,

$\Rightarrow \dfrac{AR}{18} = \dfrac{5}{7.5} \\[1em] \Rightarrow AR = \dfrac{5}{7.5} \times 18 \\[1em] \Rightarrow AR = \dfrac{1}{1.5} \times 18 \\[1em] \Rightarrow AR = 12.$

From figure,

RB = AB - AR = 18 - 12 = 6.

In right-angled ∆ABC,

⇒ AC² = AB² + BC² [By pythagoras theorem]

⇒ AC² = 18² + (7.5)²

⇒ AC² = 324 + 56.25 = 380.25

⇒ AC = $\sqrt{380.25}$ = 19.5 cm.

(i) In right-angled ∆BSR,

tan x° = $\dfrac{\text{Perpendicular}}{\text{Base}}$

= $\dfrac{RB}{RS} = \dfrac{6}{5}$.

Hence, tan x° = $\dfrac{6}{5}$.

(ii) In right-angled ∆ABC,

sin y° = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{BC}{AC}$ = $\dfrac{7.5}{19.5}$

= $\dfrac{75}{195}$ = $\dfrac{5}{13}$

Hence, sin y° = $\dfrac{5}{13}$.