Question

Find the measurement of the angles $\angle APQ$,$\angle QPR$ and $\angle BPS$.

Answer 1

$LRSP+LRQP={180}^o$ ($PQRS$ is a cyclic quadrailateral)

${65}^o+\angle RQP={180}^0$

$\angle RQP={115}^o$

In $△QPR$

$\angle RPQ={180}^o-({115}^o+{25}^o)$

$={180}^o-{140}^o$

$\angle RPQ={40}^o$

$OP\perp AB$

$\angle ROP=2\angle RSP=2\times {65}^o$

$\angle ROP={130}^o$

$\angle OQR=\angle OPR$ (angle opposite to equal side in $△OPR$)

$\angle OPR=\frac{{180}^o-{130}^o}{2}$

$\angle OPR={25}^o$=$\angle OPS={40}^o-{25}^o={15}^o$

$\angle OPB={90}^o$

$\angle BPS={90}^o-\angle OPS={90}^o-{15}^o$

$LBPS={75}^o$

$\angle APQ=\angle OPA-(\angle OPR-\angle RPQ)$

${90}^o-{25}^o-{40}^o$

$LAPQ={25}^o$