Question

Plus can anyone solve the 11th one

Answer 1

AB = AC (given)

so, ∠ACB = ∠ABC (the angles which are opposite to equal sides of a triangle are also equal with one another)

Let, ∠ABC = x°

therefore, ∠ACB = ∠ABC = x°

In ∆ABC,

∠ABC + ∠ACB + ∠BAC = 180°

or, x° + x° + 50° = 180°

or, 2x° = 130°

or, x° = 65°

So, ∠ACB = 65° and ∠ABC = 65°

Now LM is parallel to BC.

So, ∠ALM = 65° and ∠AML = 65° [CORRESPONDING ANGLES ARE ALWAYS EQUAL TO EACH OTHER]

Finally, we can say that,

∠AML + ∠CML = 180° (since the sum of all angles on the same straight line with a common vertex = 180°)

or, ∠CML = 180° - 65° = 115° [ANS.]

Answer 2

Answer:

This is a question of triangle