# Sketch an angle \theta in standard position such that the terminal side of \theta falls on the...

## Question:

Sketch an angle {eq}\theta {/eq} in standard position such that the terminal side of {eq}\theta {/eq} falls on the line {eq}3x - 4y = 0 {/eq} where {eq}x = -4 {/eq}, and then determine the 6 trigonometric functions for {eq}\theta {/eq} (correctly labeled).

## Sketching Angles by Lines:

While sketching an angle, the standard placement of the base leg is along the positive direction of the {eq}x- {/eq}axis.

The terminal leg of the angle must meet the positive {eq}x- {/eq}axis at {eq}(0,0) {/eq}.

If the terminal leg is determined by a line, there must be a specification on which part of the line is to be considered.

This is because the terminal leg is a ray and a line can be split into two rays.

## Answer and Explanation: 1

The angle lies on the line {eq}3x - 4y = 0 {/eq}.

To sketch the angle, plot the line {eq}3x - 4y = 0 {/eq}.

When {eq}x=0 {/eq}, {eq}0-4y=0 {/eq}, and consequently, {eq}y=0 {/eq}.

When {eq}x=-4 {/eq}, {eq}-12-4y=0 {/eq}, and consequently, {eq}y=-3 {/eq}.

So, the line passes through {eq}(0,0) \text{ and } (-4,-3) {/eq}.

Plot the points and then join them using a straight line to plot {eq}3x - 4y = 0 {/eq}.

The required angle is formed by the line on the point where {eq}x=-4 {/eq}.

So, the terminal leg of the angle has the point {eq}(-4,-3) {/eq}.

Since {eq}(-4,-3) {/eq} lies in quadrant III, the terminal leg of the angle also lies in quadrant III.

The angle is depicted in the graph given below:

The angle forms an acute triangle with the negative {eq}x- {/eq}axis and a line parallel to the negative {eq}y- {/eq}axis.

So, the trigonometric ratios can be used to determine the values of the trigonometric functions.

The point {eq}(-4,-3) {/eq} implies that the acute angle has a base length of 4 units and a perpendicular length of 3 units.

Use the Pythagoras theorem to determine the hypotenuse.

{eq}H = \sqrt{ 4^2+3^2} \\ H=\sqrt{ 16+9}\\ H=\sqrt{25}\\ H=5 {/eq}

So, the sine of the acute angle, which is equivalent to the ratio of perpendicular and the hypotenuse is equal to {eq}\dfrac{3}{5} {/eq}.

However, the angle {eq}\theta {/eq} lies in the third quadrant and the sine of angles in quadrant III are negative.

So, {eq}\bf{\sin \theta = \dfrac {-3}{5}} {/eq}.

Similarly, the cosine of the acute angle is {eq}\dfrac{4}{5} {/eq}.

However, the cosine of {eq}\theta {/eq} will also be negative for the same reason mentioned above.

So, {eq}\bf{\cos \theta = \dfrac{-4}{5}} {/eq}.

The remaining trigonometric values can be computed as follows:

- {eq}\begin{aligned} \tan \theta &= \dfrac {\sin \theta}{\cos \theta }\\ &= \dfrac{\dfrac{-3}{5}}{\dfrac{-4}{5}}\\ &=\dfrac{3}{4} \end{aligned} {/eq}

- {eq}\begin{aligned} \csc \theta &= \dfrac {1}{\sin \theta }\\ &= \dfrac{1}{\dfrac{-3}{5}}\\ &=\dfrac{-5}{3} \end{aligned} {/eq}

- {eq}\begin{aligned} \sec \theta &= \dfrac {1}{\cos \theta }\\ &= \dfrac{1}{\dfrac{-4}{5}}\\ &=\dfrac{-5}{4} \end{aligned} {/eq}

- {eq}\begin{aligned} \cot \theta &= \dfrac {1}{\tan \theta }\\ &= \dfrac{1}{\dfrac{3}{4}}\\ &=\dfrac{4}{3} \end{aligned} {/eq}

So, the values of the trigonometric functions for the angle are {eq}\bf{\sin \theta = \dfrac {-3}{5} , \cos \theta = \dfrac {-4}{5} , \tan \theta = \dfrac {3}{4} , \csc \theta = \dfrac {-5}{3} , \sec \theta = \dfrac {-5}{4}, \text{ and } \cot \theta = \dfrac{4}{3}} {/eq}.

#### Learn more about this topic:

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Chapter 22 / Lesson 11Learn what trigonometry is and what trigonometric functions are. Understand the examples of how to use each function, as well as know the instances when it is useful to use trigonometry.