In △ ABC, ∠ A = 100° and ∠ C = 50°. Which is its shortest sid

1.	In △ ABC, ∠ A = 100° and ∠ C = 50°. Which is its shortest side?
Answer» In △ ABC it is given that ∠ A = 100^o and ∠ C = 50^o Based on the sum property of the triangle ∠ A + ∠ B + ∠ C = 180^o To find ∠ B ∠ B = 180^o – ∠ A – ∠ C By substituting the values in the above equation ∠ B = 180^o – 100^o – 50^o By subtraction ∠ B = 180^o – 150^o ∠ B = 30^o So we get ∠ B < ∠ C < ∠ A i.e. AC < AB < BC Hence, AC is the shortest side in △ ABC.

In △ ABC, ∠ A = 100° and ∠ C = 50°. Which is its shortest side?

Answer»

In △ ABC it is given that ∠ A = 100^o and ∠ C = 50^o

Based on the sum property of the triangle

∠ A + ∠ B + ∠ C = 180^o

To find ∠ B

∠ B = 180^o – ∠ A – ∠ C

By substituting the values in the above equation

∠ B = 180^o – 100^o – 50^o

By subtraction

∠ B = 180^o – 150^o

∠ B = 30^o

So we get ∠ B < ∠ C < ∠ A

i.e. AC < AB < BC

Hence, AC is the shortest side in △ ABC.