Question

In the given figure, ABCD is a kite $(BC>AB)$ inside the circle given below, AP and CQ are the tangents to the circle at A and C respectively. If $\angle DAB:\angle DCB=11:7$, then the value of $\frac{\angle BCQ+\angle DAP}{{360}^{\circ }}$ is

• A.
  1:1
• B.
  1:2
• C.
  1:3
• D.
  1:4

Answer 1

The correct option is D

1:4
Given that ABCD is a kite with $BC>AB$ and $\angle DAB:\angle DCB=11:7$.
Let O be the centre of the circle.

$In\Delta ADCand\Delta ABCHere,AB=ADandBC=CD$ $(\because ABCDisakite)$
Also, AC is common in both triangles
$\therefore \Delta ADC\cong \Delta ABC$
(By SSS congruency rule)

$\Rightarrow \angle ACD=\angle ACB\left(ByCPCT\right)$

Let, $\angle DAB=11xand\angle DCB=7x\therefore \angle DAB:\angle DCB=11:7\therefore \angle BCA=\angle DCA=\frac{7x}{2}\left(ByCPCT\right)$

Since, CQ and AP are tangents.

$\therefore OC\perp CQandOA\perp AP$

$\Rightarrow \angle BCQ={90}^{\circ }–\frac{7x}{2}\dots ..\left(i\right)$

and $\angle CAD=\angle CAB=\frac{11x}{2}\Rightarrow \angle DAP={90}^{\circ }–\frac{11x}{2}\dots ..\left(ii\right)$

Now, in $\Delta ABC,\angle B={90}^{\circ }\left(Angleinasemicircleisarightangle\right)$

Also, $\angle CAB+\angle B+\angle BCA={180}^{\circ }\Rightarrow \frac{11x}{2}+{90}^{\circ }+\frac{7x}{2}+{180}^{\circ }\Rightarrow \frac{18x}{2}={180}^{\circ }-{90}^{\circ }={90}^{\circ }\Rightarrow x={10}^{\circ }\dots ..\left(iii\right)Consider,\angle BCQ+\angle DAPFrom\left(i\right)and\left(ii\right)={90}^{\circ }-\frac{7x}{2}+{90}^{\circ }-\frac{11x}{2}={180}^{\circ }-9x={180}^{\circ }-{90}^{\circ }........\left[from\right(iii\left)\right]={90}^{\circ }$

$\therefore \frac{\angle BCQ+\angle DAP}{{360}^{\circ }}=\frac{{90}^{\circ }}{{360}^{\circ }}=\frac{1}{4}$

Hence, the correct answer is option d.