Q29 of 93 Page 6

The sum of two angles of a triangle is 120° and their difference is 20°. Find all the three angles of the triangle.

Let three angles of the triangle be A, B and C.
Let A + B = 120° ….(i)
and A - B = 20 ….(ii)
Solving equations (i) and (ii), we get 
∠A = 70° and ∠B = 50°
Now the sum of three angles of a triangle is 180°
... ∠A + ∠B + ∠C = 180°
70° + 50° + ∠C = 180°
... ∠C = 60°
Hence, three angles of DABC are 70°, 50° and 60°.

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27 In the figure, AB and CD are parallel lines. The bisectors of the interior angles on the same side of the transversal EF intersect at P. Prove that ∠GPH = 90°.
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         31 In the figure, the bisectors of DABC and DACB intersect each other at O. 
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