Question

9. The vertical angle of an isosceles triangle is twice the sum of its base angles. Find each angle of the triangle.​

Answer 1

Step-by-step explanation:

• Hope it would be helpful !

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

• The vertical angle of an isosceles triangle is twice the sum of its base angles.

Let assume that

• Given triangle be ABC such that ∠A is vertical angle and AB = AC

We know that,

Angle opposite to equal sides of a triangle are equal.

So, it means ∠B = ∠C - -------- (1)

Further, given that

$\rm :\longmapsto\:\angle A = 2(\angle B + \angle C)$

$\rm \implies\:\angle A = 2(\angle B + \angle B)$

$\rm :\longmapsto\:\angle A = 2 \times 2\angle B$

$\bf\implies \:\boxed{\tt{ \angle A = 4\angle B \: }} - - - - (2)$

We know that,

Sum of all interior angles of a triangle is supplementary.

So,

$\rm :\longmapsto\:\angle A + \angle B + \angle C = 180 \degree \:$

$\rm :\longmapsto\:4\angle B + \angle B + \angle B = 180 \degree \:$

$\rm :\longmapsto\:6\angle B = 180 \degree \:$

$\rm \implies\:\angle B = 30\degree$

So,

$\rm \implies\:\angle C = 30\degree$

and

$\rm \implies\:\angle A = 4 \times 30\degree = 120\degree$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{\angle A = 120\degree } \\ \\ &\sf{\angle B = 30\degree }\\ \\ &\sf{\angle C = 30\degree } \end{cases}\end{gathered}\end{gathered}$

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More to Learn :-

1. Angle opposite to longest side of a triangle is always greater.

2. Side opposite to greater angle of a triangle is always longest.

3. Angle opposite to equal sides of a triangle are always equal.

4. Side opposite to equal angles of a triangle are equal.

5. Exterior angle of a triangle is equal to sum of interior opposite angles.