Question

\[\Delta ABC\] is similar to \[\Delta DEF\] by the AA similarity postulate. What is the measure of angle \[C\]?

Answer 1

Hint: From the given question we have been given asked to find the angle \[C\]. From Angle to angle similarity postulates the triangles \[\Delta ABC\] and \[\Delta DEF\] are similar to each other. From this we can say that the angle \[B\] is similar or we can say angle \[B\] is equal to angle \[E\] in the triangle \[\Delta DEF\]. Then we will use the triangle property which is that all the angles combined will form a \[{{180}^{{}^\circ }}\] and thus we find the solution to the given question. So, we proceed with the solution as follows.

Complete step-by-step solution:
In the question as we are given from angle to angle similarity postulate both the triangles are similar to each other.
From the concept of similar triangles we have that,
If two triangles are similar then the angles with respective sides are similar to each other.
So, from that we can say that the angle \[B\] is equal to angle \[E\].
\[\Rightarrow \angle \text{B=}\angle \text{E=6}{{\text{0}}^{{}^\circ }}\]
In geometry the triangle has a special property which is as follows.
All the angles in a triangle combined will form a \[{{180}^{{}^\circ }}\]. So, we get as follows.
\[\Rightarrow \angle A+\angle \text{B}+\angle C\text{=18}{{\text{0}}^{{}^\circ }}\]
\[\Rightarrow {{55}^{{}^\circ }}+6{{\text{0}}^{{}^\circ }}+\angle C\text{=18}{{\text{0}}^{{}^\circ }}\]
\[\Rightarrow {{115}^{{}^\circ }}+\angle C\text{=18}{{\text{0}}^{{}^\circ }}\]
\[\Rightarrow \angle C\text{=6}{{\text{5}}^{{}^\circ }}\]
Therefore, the \[\Rightarrow \angle C\text{=6}{{\text{5}}^{{}^\circ }}\]

Note: Students must be very careful in doing the calculations. Students must have good knowledge in the concept of similar triangles and the properties of triangles. We must know that for a triangle when all the three angles are added they are equal to \[{{180}^{{}^\circ }}\]. If we don’t know this property of a triangle it will be difficult to solve the problem.