Find the area of isosceles triangle with base 16cm and vertical angle 60° 40’

Draw CD perpendicular to AB.

∴ D is the midpoint of AB and ∠ACB = 60° 40’

Then ∠ACD = = 30° 20’ = ∠BCD

In right angled triangle ACD,

⇒ tan 30° 20’ =

⇒ tan 30° 20’ =

From tangent table, tan 30° 18’ = 0.5844 and Mean difference of 2’ = 0.0008.

∴ tan 30° 20’ = 0.5844 + 0.0008 = 0.5852

⇒ CD = 8 ÷ 0.5852 = 13.672 cm

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

⇒ Area of ΔACD = 1/2 × 8 × 13.672 = 54.688 cm²

Since ΔABC is isosceles, area of ΔACD = area of ΔBCD = 54.688 cm².

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

= 54.688 + 54.688

= 109.376 cm²

∴ The area of given isosceles triangle = 109.376 cm².