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

07.02.2018.019

This question was previously asked in
SSC CGL Previous Paper 10 (Held On: 10 August 2017 Shift 1)
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  1. 47
  2. 68
  3. 72
  4. Cannot be determined

Answer (Detailed Solution Below)

Option 2 : 68
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Detailed Solution

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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 MethodqImage664ef988a744882781856412

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°

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