Find the area of isosceles triangle with base 15cm and vertical angle 80°

Draw CD perpendicular to AB.

∴ D is the midpoint of AB and ∠ACB = 80°

Then ∠ACD = = 40° = ∠BCD

In right angled triangle ACD,

⇒ tan 40° =

⇒ tan 40° =

From tangent table, tan 40° = 0.8391.

⇒ CD = 7.5 ÷ 0.8391 = 8.938 cm

We know that Area of right angled triangle = 1/2 bh.

⇒ Area of ΔACD = 1/2 × 7.5 × 8.938 = 33.5175 cm²

Since ΔABC is isosceles, area of ΔACD = area of ΔBCD = 33.5175 cm².

∴ Area of ΔABC = Area of (ΔACD + ΔBCD)

= 33.5175 + 33.5715

= 67.035 cm²

∴ The area of given isosceles triangle = 67.035 cm².