Question

The exterior angles, obtained on producing the base of a triangle both way are 104° and 136°. Find all the angles of the triangle.

Answer 1

In the given problem, the exterior angles obtained on producing the base of a triangle both ways are 104° and 136°. So, let us draw ΔABC and extend the base BC, such that:

 ∠ACD = 104°

∠ABE = 136°

Here, we need to find all the three angles of the triangle.

Now, since BCD is a straight line, using the property, “angles forming a linear pair are supplementary”, we get

∠ACB + ∠ADC = 180°

∠ACB + 104° = 180°

∠ACB = 180° – 104°

∠ACB = 76°

Similarly, EBC is a straight line, so we get,

∠ABC + ∠ABE = 180°

∠ABC + 136° = 180°

∠ABC = 44

Further, using angle sum property in ΔABC

∠ABC + ∠ACB + ∠BAC = 180°

44 + 76 + ∠BAC = 180°

∠ABC = 180° – 120°

∠ABC = 60°

Therefore, ∠ACB = 76°, ∠BAC = 60°, ∠ABC = 44°.