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 [tex] \angle A = \angle C[/tex]
Also, AD is the angle bisector of [tex]\angle A[/tex], which implies that it cuts the angle at A in two equal halves,
let [tex] \angle A= x^{\circ}[/tex], then the bisectors cuts it in [tex](x/2)^{\circ}[/tex].
As per the given information, we know [tex]\angle ADB[/tex] is 110°, therefore, the line BDC forms a supplementary angle;
⇒[tex]\angle CDA = 180-110 =70^{\circ}[/tex]
As shown in picture given below:
By sum of all interior angles in a triangle is 180 degree, thus
[tex]x+\frac{x}{2} +70^{\circ} =180^{\circ}[/tex] or
[tex]\frac{3x}{2} = 180-70 = 110^{\circ}[/tex]
Simplify:
[tex]x=\frac{220}{3} = 73\frac{1}{3}^{\circ}[/tex]
Therefore, the [tex]\angle A =\angle C = 73\frac{1}{3}^{\circ}[/tex].
Now, to find the angle B, we have;
[tex]\angle A+ \angle B +\angle C = 180^{\circ}[/tex] [Sum of the measure of the angles in a triangle is 180 degree]
or
[tex]2\angle A+\angle B = 180^{\circ}[/tex] or
[tex]\angle B = 180^{\circ}- 2\cdot \frac{220}{3} =180^{\circ}-\frac{440}{3}[/tex]
Simplify:
[tex]\angle B = \frac{100}{3}= 33 \frac{1}{3}^{\circ}[/tex].
