Question

In isosceles triangle △ABC,

Answer 1

Answer:

Isosceles triangle:

* two sides are equal.

* the base angle are always equal and

* the altitude is a perpendicular distance from the vertex to the base.

Since, the triangle ABC is an isosceles and AC is the base

⇒ AB=BC and  $\angle A = \angle C$

Also, AD is the angle bisector of $\angle A$, which implies that it cuts the angle at A in two equal halves,

let $\angle A= x^{\circ}$, then the bisectors cuts it in $(x/2)^{\circ}$.

As per the given information, we know $\angle ADB$  is 110°, therefore, the line BDC forms a supplementary angle;

⇒$\angle CDA = 180-110 =70^{\circ}$

As shown in picture given below:

By sum of all interior angles in a triangle is 180 degree, thus

$x+\frac{x}{2} +70^{\circ} =180^{\circ}$ or

$\frac{3x}{2} = 180-70 = 110^{\circ}$

Simplify:

$x=\frac{220}{3} = 73\frac{1}{3}^{\circ}$

Therefore, the $\angle A =\angle C = 73\frac{1}{3}^{\circ}$.

Now, to find the angle B, we have;

$\angle A+ \angle B +\angle C = 180^{\circ}$    [Sum of the measure of the angles in a triangle is 180 degree]

or

$2\angle A+\angle B = 180^{\circ}$  or

$\angle B = 180^{\circ}- 2\cdot \frac{220}{3} =180^{\circ}-\frac{440}{3}$

Simplify:

$\angle B = \frac{100}{3}= 33 \frac{1}{3}^{\circ}$.