Question

In the given figure, if DABC is an isosceles triangle and ∠PAC = 110^o, find the base angle and vertex angle of the DABC.

Answer 1

Given : ∠PAC = 110°
To find :
Base angles : ∠ABC and ∠ACB
Vertex angle : ∠BAC
In quadrilateral APQC,
∠APQ + ∠PQC + ∠ACQ + ∠PAC = 360°
⇒ 90° + 90° + ∠QCA + 110° = 360°
⇒ ∠ACQ = 360° - 290°
⇒ ∠ACQ = 70°
⇒ ∠ACB = 70°    ....(i)
In ΔABC,
AB = AC                 ....(given)
⇒ ∠ACB = ∠ABC  ....(angles opposite to two equal sides are equal)
⇒ ∠ABC = 70°      ....[From (i)]
In ΔABC
∠ABC + ∠ACB + ∠BAC = 180° ....(angle sum property)
⇒ 70° + 70° + ∠BAC = 180°
⇒ ∠BAC = 180° - 140°
⇒ ∠BAC = 40°.