Answer:
[tex]\angle ABC=116^o[/tex]
Step-by-step explanation:
see the attached figure to better understand the problem
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
[tex]\angle BDA=\angle BAD=62^o[/tex] ----> the angles of the base are equals
Find the measure of the vertex angle ABD
[tex]\angle ABD=180\°-2*62\°=56\°[/tex] ------> the sum of the internal angles of a triangle is equal to 180 degrees
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 equal to 60 degrees
so
[tex]\angle CBD=60\°[/tex]
step 3
Find the measure of the angle ∠ABC
we know that
[tex]\angle ABC=\angle ABD+\angle DBC[/tex] ---> by addition angle postulate
substitute the values
[tex]\angle ABC=56^o+60^o[/tex]
[tex]\angle ABC=116^o[/tex]
