Question

In Fig. 10.139, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is
A. 52°
B. 76°
C. 156°
D. 104°

Answer 1

D. 104°

Step-by-step explanation:

Given:

Δ ABC is an isosceles triangle.

AB = AC

AB ║CD

$\angle ABC = 52\°$

We need to find the value of x.

Solution:

Δ ABC is an isosceles triangle.

AB = AC

$\angle ABC = \angle ACB = 52\°$

Now we know that;

"Sum of all angles of a triangle is 180°."

so we can say that;

$\angle ABC + \angle ACB + \angle BAC =180$

Substituting the values we get;

$52\°+52\°+\angle BAC=180\°\\\\104\°+\angle BAC=180\°\\\\\angle BAC=180\°-104\°\\\\\angle BAC=76\°$

Now Given:

AB ║CD

so we can say that;

$\angle BAC = \angle ACD$     ⇒(Alternate angles)

$\angle BAC = \angle ACD= 76\°$

Now we can say that;

$\angle ACD + \angle DCE = 180\°$  ⇒(Supplementary angles)

Substituting the values we get;

$76\°+x =180\°\\\\x = 180\° - 76\°\\\\x = 104\°$

Hence The value of x is 104°.

Answer 2

Answer:

D. 104°  

Δ ABC is an isosceles triangle.

AB = AC

AB ║CD

Solution:

Δ ABC is an isosceles triangle.

AB = AC

Now we know that;

"Sum of all angles of a triangle is 180°."

so IT CAN BE SAID;

Substituting the values we get;

Now Given:

AB ║CD

    ⇒(Alternate angles)

Now we can say that;

 ⇒(Supplementary angles)

Substituting the values we get;

Hence The value of x is 104°.