Question

$PQR$ is the tangent to a circle at $Q$ whose centre is $O$, $AB$ is a chord parallel to $PR$ and $\angle BQR={70}^0$, then $\angle$ $AQB$ is equal to

• A. ${20}^0$
• B. ${40}^0$
• C. ${35}^0$
• D. ${45}^0$

Answer 1

Answer: (B)

${40}^0$

Given- $PR$ is a tangent to the circle with centre $O$ at $Q$.

chord $AB\parallel PR$ and $\angle BQR={70}^o$.

To find out-

$\angle BQR=?$

Solution-

$AB\parallel PR$

$\therefore \angle BQR=alt\angle ABQ={70}^o$ (given)........(i)

Again $PR$ touches the circle at $Q$ and the\quad chord $BQ$ subtends $\angle BAQ$ on the cicumference at $A$ to the opposite side

$\therefore \angle BQR=\angle BAQ$ ......(alternate interior opposite angle)

$\therefore \angle BAQ={70}^o$ .......(From i)

$\therefore \angle BAQ+\angle ABQ={70}^o+{70}^o={140}^o$ ......(ii)

$\therefore \angle AQB={180}^o-(\angle ABQ+\angle BAQ)={180}^o-{140}^o={40}^o$