One of the base angles of an isosceles triangle is 40^o then  the vertex angle is?

In an isosceles triangle if a base angle is twice the vertical, calculate the angles of the triangle

In ΔABC, AC = BC. Then value of x is 

In the given figure AB = AC = AD, ∠B = 23^o and ∠D = 47^o. Then ∠BCD = 

The vertex angle of an isosceles triangle is 50^o then the base angles are?

In ΔABC, AB = AC and ∠B is more than 30^o but less than 45^o .Then m∠C cannot be

In the adjoining figure, ∠ ABC = ∠BDC = 90^o & BD = DC then the value of ∠DBC is

In the adjoining figure ∠BAC = 40^o then the value of ∠ACD  is

In an isosceles triangle if the vertex angle is twice the sum of the base angle, then the angles of the triangle are?

If AB = AC and ∠A = 50^o. Then measure of ∠ABC is