Question

In figure, $\angle ACD={120}^{\circ }$ and $\angle APB={100}^{\circ }$ ,
(i) find x and y.
(ii) If $\angle BAC=2.5y,$ find $\angle ABC$.

Answer 1

$\left(i\right)x+{100}^{\circ }={180}^{\circ }$ [Linear pair of angles]

$\Rightarrow x={80}^{\circ }$

Now, we have,

$x+y={120}^{\circ }$ [Sum of interior opposite angles is equal to the exterior angle]

$\Rightarrow {80}^{\circ }+y={120}^{\circ }$

$\Rightarrow y={40}^{\circ }$

$So,x={80}^{\circ },y={40}^{\circ }$​.

$\left(ii\right)\angle BAC=2.5y,$

$\angle BAC=2.5\times {40}^{\circ }={100}^{\circ }$

$\angle BAC+\angle ABC={120}^{\circ }$ [Sum of interior opposite angles is equal to the exterior angle]

$\Rightarrow \angle ABC={120}^{\circ }-{100}^{\circ }$

$\Rightarrow \angle ABC={20}^{\circ }$