Question

find the value of x

see attachment​

Answer 1

Answer:

34

Step-by-step explanation:

In triangle ADC,

AC = DC                   (given)

Therefore, LDAC = LACD = 56      (angles opposite to equal sides are equal)

In triangle ABD,

AD = BD                   (given)

LABD = LBAD = x               (angles opposite to equal sides are equal)

Now, by exterior angle property,

LADC = 2x

In triangle ADC,

By angle sum property,

56+56+2x=180

2x=180-112

2x=68

x=34

Answer 2

★ Solution :-

In the figure attached in the question, we are given with an triangle that dived a line which forms two different triangles. We are said that both the parts of the triangle are isosceles triangle. These type of triangles have two same sides and angles.

First, we'll find the value of the angle y which is marked in the figure attached in the answer.

Value of ∠y :-

$\sf \leadsto {Angle \: sum \: property}_{(Triangle)} = {180}^{\circ}$

$\sf \leadsto {56}^{\circ} + {56}^{\circ} + \angle{y} = {180}^{\circ}$

$\sf \leadsto {112}^{\circ} + \angle{y} = {180}^{\circ}$

$\sf \leadsto \angle{y} = 180 - 112$

$\sf \leadsto \angle{y} = {68}^{\circ}$

Now, we should find the value of the angle marked as z in the figure attached in the answer.

Value of ∠z :-

$\sf \leadsto {Straight \: line \: angle}_{(Triangle)} = {180}^{\circ}$

$\sf \leadsto \angle{y} + \angle{z} = {180}^{\circ}$

$\sf \leadsto {68}^{\circ} + \angle{z} = {180}^{\circ}$

$\sf \leadsto \angle{z} = 180 - 68$

$\sf \leadsto \angle{z} = {112}^{\circ}$

Now, let's find the value of the angle marked as x in both questions and answer's attachment.

Value of ∠x :-

$\sf \leadsto {Angle \: sum \: property}_{(Triangle)} = {180}^{\circ}$

$\sf \leadsto {112}^{\circ} + \angle{x} + \angle{x} = {180}^{\circ}$

$\sf \leadsto {112}^{\circ} + 2x = {180}^{\circ}$

$\sf \leadsto 2x = 180 - 112$

$\sf \leadsto 2x = 68$

$\sf \leadsto x = \dfrac{68}{2}$

$\sf \leadsto x = {34}^{\circ}$

Therefore, the value of ∠x is 34°.