Chapter 12.1, Problem 65E

To determine

To calculate: The sides and angle measure of the right angled triangle where $B=19°32\text{'}\text{ and }c=4.5$ .

  

Answer to Problem 65E

The triangle along with sides and acute angles is,

  

Explanation of Solution

Given information:

The right angle triangle where $B=19°32\text{'}\text{ and }c=4.5$ .

  

Formula used:

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 $\cot \theta =\frac{\text{adjacent}}{\text{opposite}}$ .

Calculation:

Consider the right angle triangle $B=19°32\text{'}\text{ and }c=4.5$ .

  

Observe that the length of hypotenuseis 4.5 units.

Label the length of opposite side of the triangle as b and adjacent side as a.

  

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 $\cot \theta =\frac{\text{adjacent}}{\text{opposite}}$ .

Apply it, to estimate the value of trigonometric ratios,

The value of cosine function is,

  $\begin{array}{c}cos19°32\text{'}=\frac{\text{adjacent}}{\text{hypotenuse}}\\ cos19°32\text{'}=\frac{a}{4.5}\\ a=\left(4.5\right)cos19°32\text{'}\\ a\approx 4.2\end{array}$

Therefore, the length of adjacent side is $4.2$ units.

The value of sine function is,

  $\begin{array}{c}\sin 19°32\text{'}=\frac{\text{opposite}}{\text{hypotenuse}}\\ \sin 19°32\text{'}=\frac{b}{4.5}\\ b=\left(4.5\right)\sin 19°32\text{'}\\ b\approx 1.5\end{array}$

Therefore, the length of opposite side is $1.5$ units.

Observe that the angle A and B are complementary angles so,

  $\begin{array}{c}\angle A+\angle B=90°\\ \angle A+19°32\text{'}=90°\\ \angle A=70°28\text{'}\end{array}$

Thus, the triangle so formed is provided below,