Question

prove that if two sides of a triangle are unequal then the angel opposite to the greater side is greater than angle opposite to smaller side​

Answer 1

Answer:

Given: △ABC, AB>AC

To Prove: ∠ACB>∠ABC

Construction: Make a point D on AB such that AD=AC and join DC

In △ACD,

AD=AC

∠ACD=∠ADC (Angles opposite to equal sides)

But ∠ADC>∠ABC (Exterior angle of the triangle is greater than the opposite interior angle)

Again, ∠ACB>∠ACD (D lies in interior of ∠ACB)

Thus, ∠ACB>∠ABC

solution