Question

Find the interior angles of the following triangles:

Answer 1

In ΔACD,
AD = CD                     ....(given)
⇒ ∠ACD = ∠CAD     ....(angles opposite to two equal sides are equal)
Now, ∠ACD = 50°     ....(given)
⇒ ∠CAD = 50°
By exterior angle property,
∠ADB = ∠ACD + ∠CAD = 50° + 50° = 100°
In ADB,
AD = BD                   ....(given)
⇒ ∠DBA =∠DAB     ....(angles opposite to two equal sides are equal)
Also, ∠ADB + ∠DBA + ∠DAB = 180°
⇒ 100 + 2∠DBA = 180°
⇒ 2∠DBA = 80°
⇒ ∠DBA = 40 °
⇒ ∠DBA = 40°
∠ BAC = ∠DAB + ∠CAD = 40° + 50° = 90°
HEnce, the interior angles of ΔABC are 50°, 90° and 40°.