Question

ABCD is a cyclic quadrilateral in which <BAD=75°, <ABD=58° and <ADC= 77°, AC and BD bisects at P. Then find <DPC.​

Answer 1

Gi᭄ven Question:-

ABCD is a cyclic quadrilateral in which ∠BAD=75°, ∠ABD=58° and ∠ADC= 77°, AC and BD bisects at P. Then find ∠DPC.

Re᭄quired Answer:-

Measure of ∠DPC is 92°.

Ex᭄planation:-

$\large \tt \red{Given:- }$

➔ABCD is a cyclic quadrilateral in which ∠BAD=75°, ∠ABD=58° and ∠ADC= 77°.

➔AC and BD bisects at P.

$\large \tt \red{To \: Find:- }$

➔∠DPC

$\large \tt \red{Cracking:- }$

➔ABCD is a cyclic quadrilateral whose vertices all lie on a single circle.

➔Angles in same segment of circle are equal, i.e.

$: \leadsto\tt{ \angle DBA= \angle DCA=58 ^\circ}$

In ΔDCA:-

$\implies \tt{ {58}^{ \circ} +{77}^{ \circ} + \angle DCA ={180}^{ \circ} }$

$\implies \tt{ {135}^{ \circ} + \angle DCA ={180}^{ \circ} }$

$\implies \tt{ \angle DCA ={180}^{ \circ} - {135}^{ \circ} }$

$\implies \tt{ \angle DCA ={45}^{ \circ} }$

So now,

$\implies \tt{ \angle PAB= \angle BAD- \angle DAC}$

$\implies \tt{\angle PAB ={75}^{ \circ} - {45}^{ \circ} }$

$\implies \tt{\angle PAB ={30}^{ \circ} }$

In ΔPAB :-

$\implies \tt{\angle PAB+\angle PBA +\angle BPA=180^\circ}$

$\implies\tt{30^\circ+58^\circ+\angle BPA=180^\circ}$

$\implies\tt{88^\circ+\angle BPA=180^\circ}$

$\implies\tt{\angle BPA=180^\circ -88^\circ }$

$\large\implies\tt{\angle BPA=92^\circ }$

$\huge { \boxed{ \underline{ \tt{ \red{So, the \: value \: of \: \angle BPA \: is \: 92^\circ. }}}}}$