Question

In the adjoining figure, AD=BD=AC; $\angle CAE={75}^0$ and $\angle ACD=x^0$, then the value of $x$ is

• A. ${45}^0$
• B. ${50}^0$
• C. ${60}^0$
• D. $37{\frac{1}{2}}^0$

Answer 1

Answer: (B)

${50}^0$

Given: $AD=BD=AC$ $\angle CAE={75}^{\circ }$
In $△ABD$
$AD=BD$
$\angle ABD=\angle BAD=x$ (Isosceles triangle property)
$\angle ADC=\angle ABD+\angle BAD=2x$ (Exterior angle property)
In $△ACD$
$AD=AC$
hence, $\angle ADC=\angle ACD=2x$ (Isosceles triangle property)
Now, $\angle CAE+\angle BAC=180$ (Linear pair)
$75+\angle BAC=180$
$\angle BAC={105}^{\circ }$
Sum of angles of triangle ABC = 180
$\angle BAC+\angle ABC+\angle ACB=180$
$105+x+2x=180$
$3x=75$
$x=25$
Thus, $\angle ACB=\angle ACD=2x=2\times 25={50}^{\circ }$