Question

in the adjoining figure x =?​

Answer 1

Given :-

A triangle ABC

To Find:-

Value of $\sf \angle x$

Solution :-

Before starting the answer , let's recall two properties of traingle

• Exterior Angle Property :- The external angle of a triangle is equal to the sum of both the angles opposite to that angle , except the angle internally making linear pair with that .

• Angle Sum Property :- The sum of all internal angles of a triangle is 180° .

• Angles opposite to equal sides are equal .

Now , let's go the question !!!!

__________________________

See attachment !!!! Here we have ;

• $\sf \angle A = 50°$

• $\sf \angle B = \angle C = y° \quad \qquad \bigg\{ \bf From \: Property \: 3 \bigg\}$

Now , By angle sum Property of triangle :-

$\quad \leadsto \quad \sf \angle A + \angle B + \angle C = 180°$

${ : \implies \quad \sf 50° + y + y = 180°}$

${ : \implies \quad \sf 2y = 180° - 50°}$

${ : \implies \quad \sf 2y = 130°}$

${ : \implies \quad \sf y = \dfrac{130°}{2}}$

${ : \implies \quad \sf y = \cancel{\dfrac{130°}{2}}}$

${ : \implies \quad \bf \therefore \quad y = 65°}$

Now , By Exterior Angle Property we have :-

$\quad \leadsto \quad \sf x = y + 50°$

${ : \implies \quad \sf x = 65° + 50°}$

${ : \implies \quad \bf \therefore \quad x = 115° }$

Henceforth , The Required Answer is 115° :)