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deff fn [24]
3 years ago
9

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

Mathematics
1 answer:
erastovalidia [21]3 years ago
8 0

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

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