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!

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=[tex]66\°[/tex] ----> the angles of the base are equals
Find the measure of the vertex angle
∠ABD=[tex]180\°-2*66\°=48\°[/tex] ------> the sum of the internal angles of a triangle is equal to [tex]180\°[/tex]
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 [tex]60\°[/tex]
so
∠CBD=[tex]60\°[/tex]
Step 3
Find the measure of the angle ∠ABC
∠ABC=∠ABD+∠DBC
substitute the values
∠ABC=[tex]48\°+66\°=114\°[/tex]
therefore
the answer is
the measure of the angle ∠ABC is [tex]114\°[/tex]
Answer:
The measure of the angle ∠ABC is 108°
Step-by-step explanation:
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
∠BDA=∠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°)
then,
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°
Find the measure of the angle ∠ABC
∠ABC=∠ABD+∠DBC
substitute the values
∠ABC= 48°+60° = 108°
Therefore
The answer is 108°
The measure of the angle ∠ABC is 108°
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