Question

In the diagram line AM and line AN are straight lines, ABC is an isosceles triangle, angle BAC=80 the bisectors of angle MBC and angle NCB meet at K. calculate angle BKC

Answer 1

Answer:

Given that

∠ABC = 40

△ABC is an isosceles triangle.

We know that in an isosceles triangle, base angles are equal.

Here, ∠BAC and ∠BCA are equal.

Let ∠BAC = ∠BCA = x

In △ABC, ∠BAC + ∠BCA + ∠ABC = 180

From given and above,

x + x + 40 = 180

2x + 40 = 180

2x = 140

x = 70

Therefore, ∠BAC = ∠BCA = 70

As

AM

and

CM

are the angle bisectors of the ∠BAC and ∠BCA respectively.

An angular bisector divides the angle into two equal halves.

Hence, ∠MAC = ∠MCA =

2

x

In △AMC, ∠MAC + ∠MCA + ∠AMC = 180

From above,

2

x

+

2

x

+ ∠AMC = 180

x + ∠AMC = 180

∠AMC = 180 − x

∠AMC = 180 − 70

∠AMC = 110

Therefore, the measure of ∠AMC is

′

110

0

′

.

Step-by-step explanation:

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