In △ABC, AB = AC = 15 cm, BC = 18 cm. Find

(i) cos ∠ABC

(ii) sin ∠ACB.

Draw AD perpendicular to BC.

D is mid-point of BC [∵ Perpendicular drawn to the unequal side of an isosceles triangle from the apex vertex bisects the side]

∴ BD = DC = 9 cm.

In right-angled triangle ABD,

⇒ AB² = AD² + BD²

⇒ 15² = AD² + 9²

⇒ AD² = (15)² - (9)²

⇒ AD² = 225 - 81

⇒ AD² = 144

⇒ AD = $\sqrt{144}$ = 12 cm.

(i) cos ∠ABC = $\dfrac{BD}{AB}$

= $\dfrac{9}{15} = \dfrac{3}{5}$.

Hence, cos ∠ABC = $\dfrac{3}{5}$.

(ii) sin ∠ACB = $\dfrac{AD}{AC}$

= $\dfrac{12}{15} = \dfrac{4}{5}$.

Hence, sin ∠ACB = $\dfrac{4}{5}$.