In the given figure, MN = RM = RP, then what is the value (in degrees) of ∠MPR?

Question

Question

This question was previously asked in

SSC CGL Previous Paper 10 (Held On: 10 August 2017 Shift 1)

Attempt in App

Answer 1

Here, ΔPRM and ΔMRN are isosceles triangles

Also, in ΔRPN

⇒ ∠RPM + ∠RNP = 102° ….(i) [EXTERIOR ANGLE PROPERTY]

⇒ ∠MRN = ∠MNR

⇒ ∠MRN + ∠MNR = ∠RMP …(ii) [EXTERIOR ANGLE PROPERTY

⇒ 2∠RNP = ∠RMP = ∠RPM

By (i) and (ii), we get

⇒ 3/2(∠RPM) = 102°

⇒ ∠RPM = 68°

∴ the correct option is 2)

Alternate Method

Let ∠RNP = x and ∠RPN = y.

Now 102° = x + y.....(i) [EXTERIOR ANGLE PROPERTY]

In triangle RNM,

2x = y …(ii) [EXTERIOR ANGLE PROPERTY

Now add both

3x = 102

x = 34.

We need to find ∠MPR = y = 2x = 2 × 34 = 68°