Question

Find the area of isosceles triangle with base 16cm and vertical angle 60 degree and 40'

Answer 1

Find answer in the image below

Answer 2

Answer:

Area of triangle = 109.36 cm²

Step-by-step explanation:

For better explanation of the solution see the attached figure :

Given : BC = 16 cm and ∠A = 60° 40'

⇒ ∠A = (60 + 40 × 0.0167)°

⇒ ∠A = 60.67°

Since, ΔABC is an isosceles triangle

⇒ AB = AC

⇒ ∠B = ∠C ( angle opposite to equal sides are equal)

Using angle sum property of triangle,

∠A + ∠B + ∠C = 180

⇒ 2∠B = 119.33

⇒ ∠B =59.665°

Now, draw a perpendicular D from A on BC

⇒ BD = BC = 8 cm

$\tan 59.665=\frac{AD}{BD}\\\\\implies AD = 13.67$

Base = 16 cm and Height = 13.67

$Area=\frac{1}{2}\times base\times height\\\\\implies Area=\frac{1}{2}\times 16\times 13.67\\\\\implies Area=109.36\text{ square cm}$