Chapter 6.2, Problem 40E

To determine

To calculate: The sides of the triangle and six trigonometric ratios of ${40}^{\circ }$ with sketch of triangle.

Answer to Problem 40E

The sketch of triangle along with sides is,

  

The six trigonometric ratios of ${40}^{\circ }$ are $\sin {40}^{\circ }=\text{0}\text{.66},cos{40}^{\circ }=0.76,\tan {40}^{\circ }=\text{0}\text{.86},\csc {40}^{\circ }=1.52,\sec {40}^{\circ }=\text{1}\text{.31}$ and $\text{ cot}{40}^{\circ }\text{ = 1}\text{.16}$ .

Explanation of Solution

Given information:

The right angle triangle with angle $\theta$ .

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 with length of its sides and angle $\theta$ .

Using protactor, sketch a right angled triangle with one of the acute angle as ${40}^{\circ }$ and measure the lengths of sides of the triangle using ruler and label the figure as,

  

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 sine function is,

  $\begin{array}{c}\sin {40}^{\circ }=\frac{\text{opposite}}{\text{hypotenuse}}\\ =\frac{\text{BC}}{\text{AC}}\\ =\frac{2.5}{3.8}\\ =0.66\end{array}$

The value of cosine function is,

  $\begin{array}{c}cos{40}^{\circ }=\frac{\text{adjacent}}{\text{hypotenuse}}\\ =\frac{\text{AB}}{\text{AC}}\\ =\frac{2.9}{3.8}\\ =0.76\end{array}$

The value of tangent function is,

  $\begin{array}{c}\tan {40}^{\circ }=\frac{\text{opposite}}{\text{adjacent}}\\ =\frac{\text{BC}}{\text{AB}}\\ =\frac{2.5}{2.9}\\ =0.86\end{array}$

The value cosecant function is,

  $\begin{array}{c}\csc {40}^{\circ }=\frac{\text{hypotenuse}}{\text{opposite}}\\ =\frac{\text{AC}}{\text{BC}}\\ =\frac{3.8}{2.5}\\ =1.52\end{array}$

The value of secant function is,

  $\begin{array}{c}\sec {40}^{\circ }=\frac{\text{hypotenuse}}{\text{adjacent}}\\ =\frac{\text{AC}}{\text{AB}}\\ =\frac{3.8}{2.9}\\ =1.31\end{array}$

The value of cotangent function is,

  $\begin{array}{c}\cot {40}^{\circ }=\frac{\text{adjacent}}{\text{opposite}}\\ =\frac{\text{AB}}{\text{AC}}\\ =\frac{2.9}{2.5}\\ =1.16\end{array}$

Thus, The six trigonometric ratios of $\theta$ are $\sin {40}^{\circ }=\text{0}\text{.66},cos{40}^{\circ }=0.76,\tan {40}^{\circ }=\text{0}\text{.86},\csc {40}^{\circ }=1.52,\sec {40}^{\circ }=\text{1}\text{.31}$ and $\text{ cot}{40}^{\circ }\text{ = 1}\text{.16}$ .