Question

In the figure, $AB=AC$ and $\angle ACD={115}^0$. Find $\angle A$

• A. ${50}^0$
• B. ${115}^0$
• C. ${65}^0$
• D. ${45}^0$

Answer 1

Answer: (A)

${50}^0$

In $△ABC$,
$\angle ACD+\angle ACB=180$ (Linear pair)
$115+\angle ACB=180$
$\angle ACB=180-115={65}^{\circ }$
Now, since, $AC=AB$
$\angle ABC=\angle ACB={65}^{\circ }$ (Isosceles triangle property)
Sum of angles of triangle = 180
$\angle BAC+\angle ABC+\angle ACB=180$
$\angle BAC+65+65=180$
$\angle BAC={50}^{\circ }$