Question

Question 24

In the given figure $\angle BAC={90}^{\circ },AD\perp BC$ and $\angle BAD={50}^{\circ }$, then $\angle ACD$ is

a) ${50}^{\circ }$
b) ${40}^{\circ }$
c) ${70}^{\circ }$
d) ${60}^{\circ }$

Answer 1

Given, $\angle BAC={90}^{\circ },AD\perp BC$ and $\angle BAD={50}^{\circ }$
In $\Delta ABD,\angle ABD+\angle DAB+\angle ADB={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow \angle ABD+{50}^{\circ }+{90}^{\circ }={180}^{\circ }\Rightarrow \angle ABD+{140}^{\circ }={180}^{\circ }\Rightarrow \angle ABD={180}^{\circ }-{140}^{\circ }\Rightarrow \angle ABD={40}^{\circ }$
Now, in $\Delta ABC,\angle A+\angle B+\angle C={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow {90}^{\circ }+{40}^{\circ }+\angle C={180}^{\circ }\Rightarrow \angle C={180}^{\circ }-{130}^{\circ }\Rightarrow \angle C={50}^{\circ }\therefore \angle ACD={50}^{\circ }$