In ∆ABC, ∠A = 90°. If AB = 7 cm and BC - AC = 1 cm, find :

(i) sin C

(ii) tan B

(a) In right ∆ABC

∠A = 90°

AB = 7 cm

BC - AC = 1 cm

⇒ BC = 1 + AC

We know that,

⇒ BC² = AB² + AC² [By pythagoras theorem]

⇒ (1 + AC)² = AB² + AC²

⇒ 1 + AC² + 2AC = 7² + AC²

⇒ 1 + AC² + 2AC = 49 + AC²

⇒ 2AC = 49 - 1

⇒ 2AC = 48

⇒ AC = $\dfrac{48}{2}$ = 24 cm.

∴ BC = 1 + AC = 25 cm.

(i) sin C = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{AB}{BC} = \dfrac{7}{25}$

Hence, sin C = $\dfrac{7}{25}$.

(ii) tan B = $\dfrac{\text{Perpendicular}}{\text{Base}}$

= $\dfrac{AC}{AB} = \dfrac{24}{7}$

Hence, tan B = $\dfrac{24}{7}$.