Question

In the given figure, Tr || VS, TU || RS, $\angle RPQ$ is twice of $\angle UVS$, then find a + b

• A. ${165}^{\circ }$
• B. ${185}^{\circ }$
• C. ${225}^{\circ }$
• D. ${155}^{\circ }$

Answer 1

The correct option is D ${155}^{\circ }$

In the given figure, Tr || VS, TU || RS, RPQ is twice of UVS
Then $\angle$RPQ=2$\angle$UVS=2(25)=50${}^0$
TU||RS then $\angle$UTQ=$\angle$SRQ=90${}^0$
And $\angle$QRP=180-$\angle$SRQ=180-90=90${}^0$
IN triangle PQR $\angle$a=180-$\angle$QRP-$\angle$RPQ=180-90-50=40${}^0$
TR||VS Then $\angle$b=$\angle$RTV+$\angle$UVS=90+25=115${}^0$
Then a+b=40+115=155${}^0$