Question

Please solve it fast......​

Answer 1

$\LARGE{ \underline{\underline{ \sf{Required \: answer:}}}}$

GiveN:

• ABCD is a cyclic quadrilateral.
• ∠BAD = 75°
• ∠ABD = 58°
• ∠ADC = 77°
• AC and BD intersect at P.

To FinD:

• ∠DPC = ?

Step-by-Step Explanation:

The opposite angles of a cyclic quadrilateral add upto 180°. That means,

⇒ ∠BAD + ∠DCB = 180°

⇒ ∠DCB + 75° = 180°

⇒ ∠DCB = 105°

And looking the other pairs of opposite angles,

⇒ ∠ADC + ∠ABC = 180°

⇒ 77° + ∠ABC = 180°

⇒ ∠ABC = 103°

If we take a look on ∠ABC then,

⇒ ∠ABD + ∠DBC = ∠ABC

⇒ 58° + ∠DBC = 103°

⇒ ∠DBC = 45°

Now in ∆DCB,

By Angle sum property of triangle,

⇒ ∠CDB + ∠DBC + ∠DCB = 180°

⇒ ∠CDB + 45° + 105° = 180°

⇒ ∠CDB + 150° = 180°

⇒ ∠CDB = 30°

⇒ Then, ∠ADB = 77° - 30° = 47°

Sim.∠ACB = 47° (∠s in same segment)

We know that,

⇒ ∠DCP + ∠ACB = ∠DCB

⇒ ∠DCP+ 47° = 105°

⇒ ∠DCP = 58°

Finally in ∆DPC,

Again by Angle sum property,

⇒ ∠CDP + ∠DPC + ∠DCP = 180°

⇒ 30° + ∠DPC + 58° = 180°

⇒ ∠DPC + 88° = 180°

⇒ ∠DPC = 92°

Thus the required measure of ∠DPC = 92° (Ans)

Answer 2

Answer:

✯ Given :-

• ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58°, and ∠ADC = 77°, AC and BD intersect at P.

✯ To Find :-

• What is the value of ∠DPC.

✯ Solution :-

➙ Since, angles in the same segment of a circle are equal,

$\therefore$ ∠DBA = ∠DCA = 58°

➣ In ∆DCA,

➙ Since, sum of the angles of a triangle is 180°,

⇒ ∠DCA = ∠CDA + ∠DAC = 180°

⇒ 58° + 77° + ∠DAC = 180°

⇒ ∠DAC = 180° - 58° - 77°

⇒ ∠DAC = 45°

Now,

⇒ ∠PAB = ∠BAD - ∠DAC

⇒ ∠PAB = 75° - 45°

⇒ ∠PAB = 30°

➣ In ∆PAB

➙ Since, sum of the angles of the triangle is 180°,

⇒ ∠PAB + ∠PBA + ∠BPA = 180°

⇒ 30° + 58° + ∠BPA = 180°

⇒ 88° + ∠BPA = 180°

⇒ ∠BPA = 180° - 88°

⇒ ∠BPA = 92°

➥ ∠BPA = ∠DPC = 92°

$\therefore$ The value of ∠DPC = $\sf\boxed{\bold{\large{92°}}}$

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