Chapter 6.2, Problem 66E

To determine

To explain: The properties of similar triangles. Also define the trigonometric ratios with the help of the properties.

Answer to Problem 66E

Two triangles are similar if their corresponding angles are congruent and all their corresponding sides are in same proportion.

Explanation of Solution

Given information:

Two similar triangles.

Construct two triangles $△PQR$ and $△STU$ . These two triangles are similar triangles that is $△PQR\sim △STU$ .

  

The corresponding angles in both the triangles $△PQR$ and $△STU$ are congruent that is,

  $\angle P=\angle S,\angle Q=\angle T\text{ and }\angle R=\angle U$ .

The corresponding sides in both the triangles $△PQR$ and $△STU$ are in same proportion that is,

  $\frac{PQ}{ST}=\frac{QR}{TU}=\frac{PR}{SU}$

Recall that the trigonometric ratios for a right angle triangle are defined as,

  $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}},cos\theta =\frac{\text{adjacent}}{\text{hypotenuse}},\tan \theta =\frac{\text{opposite}}{\text{adjacent}},\csc \theta =\frac{\text{hypotenuse}}{\text{opposite}},\sec \theta =\frac{\text{hypotenuse}}{\text{adjacent}}$ and $\cot \theta =\frac{\text{adjacent}}{\text{opposite}}$ .

Since, $\frac{PQ}{ST}=\frac{PR}{SU}$ this implies that $\frac{PQ}{PR}=\frac{ST}{SU}$ . The value of sine function is,

  $\sin R=\frac{\text{opposite}}{\text{hypotenuse}}$ and $\sin U=\frac{\text{opposite}}{\text{hypotenuse}}$ .

Therefore,

  $\begin{array}{c}\frac{PQ}{PR}=\frac{ST}{SU}\\ \sin R=\sin U\end{array}$

Since, $\frac{QR}{TU}=\frac{PR}{SU}$ this implies that $\frac{QR}{PR}=\frac{TU}{SU}$ . The value of cosine function is,

  $\cos R=\frac{\text{adjacent}}{\text{hypotenuse}}$ and $\cos U=\frac{\text{adjacent}}{\text{hypotenuse}}$ .

Therefore,

  $\begin{array}{c}\frac{QR}{PR}=\frac{TU}{SU}\\ \cos R=\cos U\end{array}$

Since, $\frac{PQ}{ST}=\frac{QR}{TU}$ this implies that $\frac{PQ}{QR}=\frac{ST}{TU}$ . The value of tangent function is,

  $\tan R=\frac{\text{opposite}}{\text{adjacent}}$ and $\tan U=\frac{\text{opposite}}{\text{adjacent}}$ .

Therefore,

  $\begin{array}{c}\frac{PQ}{QR}=\frac{ST}{TU}\\ \tan R=\tan U\end{array}$

Thus, the trigonometric ratios can be estimated regardless of the size of triangle.