1.

In a ∆ABC, ∠C = 3∠B = 2(∠A + ∠B). Find the measure of each one of ∠A, ∠B and ∠C.

Answer»

In a ∆ABC, 

∠C = 3∠B = 2(∠A + ∠B). 

∴ 3∠B = 2(∠A + ∠B) 

⇒ 3∠B = 2∠A + 2∠B 

⇒ 3∠B – 2∠B = 2∠A 

⇒ ∠B = 2∠A. 

And ∠C = 3∠B = 6∠A. (∵∠B = 2∠A) 

We know that sum of angles in a triangle is 180°. 

∴ ∠A + ∠B +∠C = 180° 

⇒ ∠A + 2∠A + 6∠A = 180° 

⇒ 9∠A = 180° 

⇒ ∠A = \(\frac{180°}{9}\)= 20°. 

Now, ∠B = 2∠A = 2× 20° = 40°. 

And ∠C = 6∠A = 6 ×20° = 120°. 

Hence, ∠A = 20°, ∠B = 40° and ∠C = 120°.



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