Question

The base angle of an isosceles triangle is ${15}^{\circ }$ more than its vertical angle. Find its each angle

Answer 1

Step: Find all angles of triangle

Let the vertical angle be = x

∴ Each base angle = x + ${15}^{\circ }$

Now, x + (x + ${15}^{\circ }$) + ( x + ${15}^{\circ }$ ) = ${180}^{\circ }$

[∵ Angle sum property of a triangle]

3x + ${30}^{\circ }$ = ${180}^{\circ }$

3x = ${180}^{\circ }$ - ${30}^{\circ }$

3x = ${150}^{\circ }$

x = ${50}^{\circ }$

∴ x + ${15}^{\circ }$ = ${65}^{\circ }$

Hence, the angles of the triangles are ${50}^{\circ }$, ${65}^{\circ }$ and ${65}^{\circ }$.