Question

In the given above figure, AB || QR, $\angle BAQ={142}^o$ and $\angle ABP={100}^o$. Find $\angle APB$.

Answer 1

$\angle BAQ+\angle BAP={180}^o$ {Linear pair or angles on the straight line}

${142}^o+\angle BAP={180}^o$

$\angle BAP={180}^o-{142}^o={38}^o$

Also,

$\angle APB+\angle ABP+\angle BAP={180}^o$ {angle sum property of a triangle ABP}

$\angle APB+{100}^o+{38}^o={180}^o$

$\angle APB={180}^o-{100}^o-{38}^o$

$\angle APB={42}^o$