PreCal Honors - Unit Test #5

PreCalculus Honors

Unit Test #5

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Part 1: Calculator Allowed

Give two angles that are coterminal with $\frac{11 π}{18}$ .

(3)

$\frac{11 π}{18} - 2 π = \frac{11 π}{18} - \frac{36 π}{18} = - \frac{25 π}{18}$ , or $\frac{11 π}{18} + 2 π = \frac{11 π}{18} + \frac{36 π}{18} = \frac{47 π}{18}$

Evaluate: $cos (10)$

(2)

$cos (10) \approx - 0.8391$ (remember that you should be in radians)

Evaluate: ${sin}^{- 1} (\frac{2}{3})$

(2)

${sin}^{- 1} (\frac{2}{3}) \approx 0.7297$ (this should be in radians as well)

The angle of inclination of a mountain with triple black diamond ski trails is 65°. If a skier at the top of the mountain is at an elevation of 4000 feet, how long is the ski run from the top to the base of the mountain?

(5)

Perhaps a picture?

An Image

An equation to solve this would be $sin 65 ° = \frac{4000}{x}$ … $x = \frac{4000}{sin 65 °} \approx 4413.5117$ .

The ski run is 4413.5117 feet long.

What is the angle of inclination required to observe a drone flying at an altitude of 40 feet when its distance from the observer is 250 feet?

(5)

Another picture?

An Image

The equation: $sin θ = \frac{40}{250} \Rightarrow θ = {sin}^{- 1} (\frac{40}{250}) \approx 9.2069 °$

PreCalculus Honors

Unit Test #5

Part 2: No Calculator Allowed

An Image

Find the value of the trigonometric function indicated above.

(4)

First, the third side: $5^{2} + 12^{2} = h^{2} \Rightarrow 25 + 144 = h^{2} \Rightarrow 169 = h^{2} \Rightarrow 13 = h$ . Since secant is hypotenuse over adjacent, $sec θ = \frac{13}{12}$ .

Convert to degrees: $\frac{7 π}{15}$

(4)

$\frac{d}{180 °} = \frac{\frac{7 π}{15}}{π} \Rightarrow \frac{d}{180 °} = \frac{7}{15} \Rightarrow d = (\frac{7 \cdot 180}{15}) ° = (7 \cdot 12) ° = 84 °$

Convert to radians: 105°

(4)

$\frac{105 °}{180 °} = \frac{r}{π} \Rightarrow r = \frac{105 π}{180} = \frac{7 π}{12}$

Evaluate: $sin (\frac{7 π}{6})$

(2)

$- \frac{1}{2}$

10.

Evaluate: $cos (\frac{5 π}{4})$

(2)

$- \frac{1}{\sqrt{2}}$

11.

Evaluate: $tan (\frac{4 π}{3})$

(2)

$\sqrt{3}$

12.

Find $csc θ$ , given that the terminal side of $θ$ passes through the point $(3, 5)$ .

(4)

First off, the distance to the point: $3^{2} + 5^{2} = r^{2} \Rightarrow 9 + 25 = r^{2} \Rightarrow 34 = r^{2} \Rightarrow r = \sqrt{34}$ . Since $csc θ = \frac{y}{r}$ , $csc θ = \frac{\sqrt{34}}{5}$ .

13.

Find the values of all other trigonometric functions if $tan θ = - \frac{5}{12}$ and $cos θ < 0$ .

(6)

Since $cos θ < 0$ and $tan θ < 0$ makes $sin θ > 0$ (Quadrant 2). ${(- 12)}^{2} + 5^{2} = r^{2} \Rightarrow 144 + 25 = r^{2} \Rightarrow 169 = r^{2} \Rightarrow 13 = r$ . We now have $x = - 12$ , $y = 5$ and $r = 13$ . $sin θ = \frac{5}{13}$ , $cos θ = - \frac{12}{13}$ , $sec θ = - \frac{13}{12}$ , $csc θ = \frac{13}{5}$ and $cot θ = - \frac{12}{5}$ .

Page last updated 09:08 2021-04-22