Question

Match the two column;
$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{(A) In the given figure,}& \text{(p)}{60}^{\circ }\\ \text{ABCD is a cyclic Quadrilateral.}& \\ \text{O is the centre of the circle.}& \\ \text{If}\angle \text{BOD}={160}^{\circ }\text{find the Measure of}\angle \text{BPD.}& \end{array}$


$\begin{array}{cc}\text{(B) In given figure, ABCD is a cyclic}& \text{(q)}{65}^{\circ }\\ \text{quadrilateral whose side AB is a diameter}& \\ \text{of the circle through A, B, C, D.}& \\ \text{If}\angle {\text{ADC = 130}}^{\circ }\text{, find}\angle \text{BAC.}& \end{array}$


$\begin{array}{cc}\text{(C) In the given figure BD = DC and}& {\text{(r) 40}}^{\circ }\\ \angle {\text{CBD = 30}}^{\circ }\text{Find m}\left(\angle \text{BAC}\right)& \end{array}$


$\begin{array}{cc}\text{(D) In the given figure, O is the centre of the}& {\text{(s) 100}}^{\circ }\\ {\text{arc ABC subtends an angle of 130}}^{\circ }\text{at the centre.}& \\ \text{If AB is extended to Pm find}\angle \text{PBC.}& \end{array}$


Choose the correct option:

• A. A-s; B-r; C-p; D-q
• B. A-s; B-p; C-q; D-r
• C. A-p; B-q; C-r; D-s
• D. A-s; B-p; C-r; D-q

Answer 1

The correct option is A

A-s; B-r; C-p; D-q

(a) $\left(A\right)\to s;\left(B\right)\to r;\left(C\right)\to p,\left(D\right)\to q$
(A) Consider the arc BCD of the circle. This arc makes angle $\angle BOD={160}^{\circ }$ at the centre of the circle and $\angle BAD$ at a point A on the circumference.



$\therefore \angle BAD=\frac{1}{2}\angle BOD={80}^{\circ }$
Now, ABPD is a cyclic quadrilateral.
$\Rightarrow \angle BAD+\angle BPD={180}^{\circ }\Rightarrow {80}^{\circ }+\angle BPD={180}^{\circ }\Rightarrow \angle BPD={100}^{\circ }$

(B) Since ABCD is a cyclic quadrilateral.
$\therefore \angle ADC+\angle ABC={180}^{\circ }\Rightarrow {130}^{\circ }+\angle ABC={180}^{\circ }\Rightarrow \angle ABC={50}^{\circ }$



Since $\angle ACB$ is the angle in a semicircle.
$\therefore \angle ACB={90}^{\circ }$
Now, in $\Delta ABC$, we have
$\angle BAC+\angle ACB+\angle ABC={180}^{\circ }\Rightarrow \angle BAC={90}^{\circ }+{50}^{\circ }={180}^{\circ }\Rightarrow \angle BAC={40}^{\circ }$

(C) BD=DC
$\angle BCD=\angle CBD={30}^{\circ }$
So, $\angle BDC={120}^{\circ }$
As ABCD is cyclic
$\therefore \angle BAC={60}^{\circ }$

(D) $\angle PBC={65}^{\circ }$ (If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite opposite angle)