In triangle ABC; ∠ABC = 90°, ∠CAB = x°, $\text{tan x°} = \dfrac{3}{4}$ and BC = 15 cm. Find the measures of AB and AC.

Given:

$\text{tan x°} = \dfrac{3}{4}\\[1em] \text{tan x°} = \dfrac{Perpendicular}{Base} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{BC}{AB} = \dfrac{3}{4}$

∴ If length of AB = 4x unit, length of BC = 3x unit.

BC = 15 cm (∵ Given)

∴ 3x = 15

⇒ x = $\dfrac{15}{3}$ = 5 cm

∴ AB = 4x = 4 x 5 = 20 cm

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AB is hypotenuse)

⇒ AC² = (15)² + (20)²

⇒ AC² = 225 + 400

⇒ AC² = 625

⇒ AC = $\sqrt{625}$

⇒ AC = 25 cm

Hence, AB = 20 cm and AC = 25 cm.