One base angle of the isosceles ΔABC is twice of the vertical angle. Let’s write the measurement of three of the triangle.

Given: isosceles ΔABC and one base angle is twice of the vertical angle

To find the measurement of three of the triangle

The figure for the above question is as shown below,

Now given ΔABC is an isosceles triangle,

AB = AC and ∠ABC = ∠ACB……..(i)

Now in ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ABC + ∠ACB = 180°

Substituting equation (i) in above equation we get

∠BAC + ∠ABC + ∠ABC = 180°

∠BAC + 2∠ABC = 180°………..(ii)

In this triangle ∠ ABC and ∠ACB are base angles

It is given base angle is twice of the vertical angle

⇒ ∠ABC = 2∠BAC…….(iii)

Substituting equation (iii) in equation (ii) we get

∠BAC + 2(2∠BAC) = 180°

⇒ ∠BAC + 4∠BAC = 180°

⇒ 5∠BAC = 180°

⇒ ∠BAC = 36°……(iv)

Substituting equation (iv) in equation (iii) we get

⇒ ∠ABC = 2(36°)

⇒ ∠ABC = 72°…….(v)

Substituting equation (v) in equation (i) we get

∠ACB = 72°…..(vi)

So from equation (iv), (v) and (vi), the measurements of the angles

of ΔABC are

∠BAC = 36°, ∠ABC = 72°, ∠ACB = 72°