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Question
In a triangle, the sum of two angles is 139° and their difference is 5°; find each angle of the triangle.
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Solution
Let ABC be a triangle such that
∠A + ∠B = 139° ....(i)
and ∠A - ∠B = 5° ....(ii)
Adding (i) and (ii), we get
2∠A = 144°
⇒ ∠A = 72°
From (i), we have
∠B = 139° - 72° = 67°
Now, 3rd angle
= 180° - (∠A + ∠B)
= 180° - 139°
= 41°
Thus, the angles of a triangle are 72°, 67° and 41°.
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