Question

An equilateral triangle and an isosceles triangle share a common side. What is the measure of angle ABC? I need help with question 13 please asap!

Answer 1

Step 1

Find the measure of the vertex angle ∠ABD of an isosceles triangle

we know that

An isosceles triangle has two equal sides and two equal angles

In this problem

∠ABD=∠BAD=$66\°$ ----> the angles of the base are equals

Find the measure of the vertex angle

∠ABD=$180\°-2*66\°=48\°$ ------> the sum of the internal angles of a triangle is equal to $180\°$

Step 2

Find the measure of the angle  ∠CBD in the equilateral triangle

we know that

A equilateral triangle has three  equal sides and three equal angles

The measure of the internal angle in a equilateral triangle is $60\°$

so

∠CBD=$60\°$

Step 3

Find the measure of the angle ∠ABC

∠ABC=∠ABD+∠DBC

substitute the values

∠ABC=$48\°+66\°=114\°$

therefore

the answer is

the measure of the angle ∠ABC is $114\°$

Answer 2

The answer is ABC equals 48.
That is becuase two sides of the triangle are 66.
66+66=132
180-132=48