Question

in triangle pqr PQ = QR and angle PRS(exterior angle) =100°.
Fing angle QPR.​

Answer 1

SOLUTION

Given,

In ∆PQR

PQ= QR (given)

angle PRS (exterior angle= 100°)

=) Let PQ= angle x & similarly QR= x

=) angle PRS+ angle PQS= 180°

=) 100°+angle PQS 180°

=) PQS= (180-100)°

=) angle PQS= 80°

=) angle PQS= 80°

Then similar as angle PQS= angle QPR

So, angle QPR = 80°

Hope it helps ☺️

Answer 2

Que. in triangle pqr PQ = QR and angle PRS(exterior angle) =100°.

Fing angle QPR.

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As per the information given in the question

PQ = QR

PRS(exterior angle) =100°

Accordin to question we have to find angle QPR

The exterior angle PRS = 100°

The exterior angle PRS and the interior angle PRQ are on straight line.

the sum of angles on a straight line is equal to 180°. So,

angle PRS + angle PRQ = 180°

100° + angle PRQ = 180°

angle PRQ = 180° - 100°

angle PRQ = 80°.

As per the information given in the question PQ = QR

So the given ∆PQR is isosceles.

As per the property of isosceles triangle the angles opposite to equal sides are also equal to each other.

angle PRQ = angle QPR So,

angle QPR = 80°