Question

This is a geometry question Ive been stuck with can you please help me? Im offering a hundred points for ho ever can help me. I really do need help​

Answer 1

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Answer 2

Step-by-step explanation:

Given:

<CAB = 58°

<FEB = 42°

<EFB = 65°

To find:

p° and q°

Construction:

Extend AB to D.

Solution:

For angle p°,

In triangle EFB

<F = 65°

<E = 42°

< B = ?

Using angle sum property of a triangle,

<EFB+<FBE+<BEF = 180°

65°+<FBE+42° = 180°

<FBE+107° = 180°

<FBE = 180°-107°

<FBE = 73°

And, we can see that AD is a straight line

So,

<ABC+<CBD = 180°

73°+<CBD = 180°

<CBD = 180°-73°

<CBD = 107°

p° = 107°

For angle q°,

In triangle ABC,

<A = 58°

<B = 73°

<C = ?

Now,

Using angle sum property of a triangle,

<ABC+<BCA+<CAB = 180°

73°+<BCA+58° = 180°

<BCA+131° = 180°

<BCA = 180°-131°

<BCA = 49°

q° = 49°

Hence, p° = 107° and q° = 49°.

Solution is very lengthy but, easy also.....So, don't worry about it.

♥️Thanks buddy......!!♥️