Question

In ∆PQR, PQ=PR . The bisector of angle Q and angle R meet at O . If angle P= 70° .find angle angle QOR​

Answer 1

If we draw a circle outside the triangle it will look like the attachment .

• PQ=PR
• <P=70°

Now

$\\ \sf\longmapsto 2<QOR=<P$

$\\ \sf\longmapsto <QOR=\dfrac{<P}{2}$

$\\ \sf\longmapsto <QOR=\dfrac{70}{2}$

$\\ \sf\longmapsto <QOR=35°$

Answer 2

Answer:

125 degrees

Step-by-step explanation:

As PR=PQ angle opposite to equal sides are equal so angle R=angle Q

angle R +angle Q=180-70 (angle sum property)

angle R +angle Q=110

2angle R=110

angle R=110/2=55 degrees

angle ORQ=55/2=27.5 degrees (bisectors)

angle OQR=55/2=27.5 degrees

angle QOR=180-(27.5+27.5) (angle sum property)

                  =180-55=125 degrees