In the given figure, ABCD is a cyclic quadrilateral in which =75^ , =58^ and =77^ , AC and BD intersect at P. The measure of is:

In the given figure, ABCD is a cyclic quadrilateral in which =75^…

MCQ

In the given figure, $ABCD$ is a cyclic quadrilateral in which $\angle\text{BAD}=75^\circ,$ $\angle\text{ABD}=58^\circ$ and $\angle\text{ADC}=77^\circ, AC$ and $BD$ intersect at $P.$ The measure of $\angle\text{DPC}$ is:

A
$94^\circ$
B
$105^\circ$
✓
$92^\circ$
D
$90^\circ$

✓

Answer

Correct option: C.

$92^\circ$

Since $AD$ acts as a chord also, So, $\angle\text{ABD}=\angle\text{ACD}=58^\circ$
Again as $CD$ also acts as a chord also, therefore,
$\angle\text{DBC}=\angle\text{DAC}$
Now, $\angle\text{ABC}=\angle\text{ABD}+\angle\text{DBC}$
Also, $\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-77^\circ=103^\circ$
And therefore
$\angle\text{DBC}=103^\circ-58^\circ=45^\circ$
Hence, $\angle\text{DAC}=45^\circ$
Since,
$\angle\text{DAC}=45^\circ$
So, $\angle\text{CAB}=75^\circ-45^\circ=30^\circ$
But, $\angle\text{CAB}=\angle\text{BDC}$
$\Rightarrow\angle\text{BDC}=30^\circ$
Now, In triangle $CPD,$
$\angle\text{C}+\angle\text{P}+\angle\text{D}=180^\circ$
$\Rightarrow58^\circ+\angle\text{P}+30^\circ=180^\circ$
$\Rightarrow\angle\text{P}=180^\circ-30^\circ-58^\circ=92^\circ$