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Homework 2
Section 7.1 #24) Find the arc length indicated in the following image:

Solution: Recall the fundamental relationship:

$s=r\theta$ , where

$s$ denotes arc length,

$r$ denotes the radius, and

$\theta$ the (radian) measure of the angle. In this problem, we are told

$r=4.5$ and

$\theta=\dfrac{2\pi}{5}$ and are asked to find

$s$ . Therefore

$s = (4.5)\dfrac{2\pi}{5} = \dfrac{9\pi}{5} \approx 5.65 \mathrm{cm}.$

Section 7.1 #27) Convert to degree measure:

$\dfrac{\pi}{9}$ radians.
Solution: Recall that

$180^{\circ}=\pi \hspace{2pt} \mathrm{radians}$ so that

$\dfrac{180^{\circ}}{\pi \hspace{2pt} \mathrm{radians}} = 1$ . Therefore compute

$\begin{array}{ll} \dfrac{\pi}{9} \hspace{2pt} \mathrm{radians} &= \left( \dfrac{\pi}{9} \hspace{2pt} \mathrm{radians} \right) (1) \\ &= \left( \dfrac{\pi}{9} \hspace{2pt} \mathrm{radians} \right) \left( \dfrac{180^{\circ}}{\pi \hspace{2pt} \mathrm{radians}} \right) \\ &= \left( \dfrac{180}{9} \right)^{\circ} \\ &= 20^{\circ}. \end{array}$

Section 7.1 #28) Convert

$-\dfrac{5\pi}{4}$ radians to degrees.
Solution: Calculate

$-\dfrac{5\pi}{4} \mathrm{\hspace{2pt} rad} = \left( -\dfrac{5\pi}{4} \mathrm{\hspace{2pt} rad} \right)\left( \dfrac{360^{\circ}}{2\pi \mathrm{\hspace{2pt} rad}} \right) = -\left( \dfrac{5(360)}{2(4)} \right)^{\circ}=-225^{\circ}.$

Section 7.1 #35) Convert

$-540^{\circ}$ to radians.
Solution: Calculate

$-540^{\circ}= \left( -540^{\circ} \right) \left( \dfrac{2\pi \mathrm{\hspace{2pt} rad}}{360^{\circ}} \right)=-\dfrac{540(2\pi)}{360} \mathrm{\hspace{2pt} rad}=-3\pi \mathrm{\hspace{2pt} rad}.$

Section 7.1 #43) Find the length of the arc of a circle of radius

$10$ centimeters subtended by the central angle of

$50^{\circ}$ .
Solution: Recall the fundamental relationship

$s=r\theta$ (

$\theta$ must be in radians!!!!). In this problem we are told

$r=10$ and

$\theta=50^{\circ}$ . First convert

$50^{\circ}$ into radians:

$50^{\circ} = \left( 50^{\circ} \right) \left( \dfrac{\pi \hspace{2pt} \mathrm{radians}}{180^{\circ}} \right) = \dfrac{5\pi}{18} \mathrm{radians}.$ With these values, we see that the desired arc length,

$s$ , is

$s = 10 \left( \dfrac{5\pi}{18} \right) = \dfrac{50\pi}{18} \approx 8.73.$

Section 7.2 #12: Find the missing sides if side

$a$ is opposide angle

$A$ , side

$b$ is opposite angle

$B$ , and side

$c$ is the hypotenuse:

$\tan A = \dfrac{5}{12}, b=6.$ Solution: First draw the described triangle:

By definition,

$\tan(A)=\dfrac{a}{b}=\dfrac{a}{6}$ by what we are told. However we are also told that

$\tan(A) = \dfrac{5}{12}$ . To find

$a$ equate these two to see that

$\dfrac{5}{12} = \tan(A) = \dfrac{a}{6},$ and solve for

$a$ by multiplying by

$6$ to get

$a = 6\dfrac{5}{12} = \dfrac{5}{2}$ . To find the hypotenuse

$c$ , the Pythagorean theorem tells us that

$a^2+b^2=c^2,$ so plug in

$a$ and

$b$ to get

$\left( \dfrac{5}{2} \right)^2 + 6^2 = c^2,$ and simplify the left-hand side to get

$\dfrac{25}{4} + 36 = c^2,$ or equivalently

$\dfrac{25+144}{4} = c^2,$ or

$\dfrac{169}{4}=c^2,$ so that the hypotenuse is

$c = \sqrt{ \dfrac{169}{4} } = \dfrac{\sqrt{169}}{\sqrt{4}} = \dfrac{13}{2}.$ (note: we ignored the negative solution because we are solving for the leg of a triangle, a length, which cannot be negative!)

Section 7.2 #17: Find

$\sin(A)$ in the following picture:

Solution: To find

$\sin(A)$ we need to know the length of the hypotenuse. To find that, use the Pythagorean theorem to write

$10^2+4^2=\mathrm{hyp}^2.$ Simplifying on the left yields

$116=\mathrm{hyp}^2$ , so

$\mathrm{hyp}=\sqrt{116}$ (we ignore the negative solution to this equation because it is not physically meaningful). Therefore we may now compute

$\sin(A) = \dfrac{\mathrm{opposite \hspace{2pt} of \hspace{2pt}} A}{\mathrm{hypotenuse}} = \dfrac{10}{\sqrt{116}}.$

Section 7.2 #31: Solve for the unknown sides of the following triangle:

Solution:
Solve for side opposite the

$45^{\circ}$ angle
Call this side

$b$ . We see that

$\sin(45^{\circ})=\dfrac{b}{15\sqrt{2}}$ and so solving for

$b$ yields

$b=15\sqrt{2}\sin(45^{\circ})$ .

Solve for side opposite of angle

$A$
Call this side

$a$ . We see that

$\cos(45^{\circ})=\dfrac{a}{15\sqrt{2}}$ and so solving this for

$a$ yields

$a=15\sqrt{2}\cos(45^{\circ})$ .

Section 7.2 #51: There is a lightning rod on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be

$36^{\circ}$ . From the same location, the angle of elevation to the top of the lightning rod is measured to be

$38^{\circ}$ . Find the height of the lightning rod.
Solution: Draw this problem, labelling the height of the building as

$h_B$ and the height of the lightning rod as

$h_L$ :

There are two natural triangles to work with that appear in the image:

and

In the first of those two triangles, we use the tangent function to write

$\tan(38^{\circ}) = \dfrac{h_B+h_L}{500}.$ Solving for

$h_B+h_L$ yields

$h_B+h_L=500\tan(38^{\circ})$ . In the second of the two triangles, we use the tangent function to write

$\tan(36^{\circ})=\dfrac{h_B}{500}.$ Solving for

$h_B$ yields

$h_B=500\tan(36^{\circ})$ . We would like to solve for

$h_L$ . Since

$(h_B+h_L)-h_B=h_L$ , this shows us that

$h_L = (h_B+h_L)-h_b = 500\tan(38^{\circ}) - 500\tan(36^{\circ}) \approx 27.37 \mathrm{\hspace{2pt} feet}.$