Question

An equilateral triangle and an isosceles triangle share a common side. what is the measure of ABC ?

Answer 1

Answer:

$\angle ABC=116^o$

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

$\angle BDA=\angle BAD=62^o$ ----> the angles of the base are equals

Find the measure of the vertex angle ABD

$\angle ABD=180\°-2*62\°=56\°$ ------> 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

$\angle CBD=60\°$

step 3

Find the measure of the angle ∠ABC

we know that

$\angle ABC=\angle ABD+\angle DBC$ ---> by addition angle postulate

substitute the values

$\angle ABC=56^o+60^o$

$\angle ABC=116^o$