Question

what do they call the big gra
on an orbiting satellite?
for the first eight exercises, find the length x. for the remaining exercises, find the length n
each answer to the nearest tenth. cross out each box that contains a correct answer. wher
the remaining boxes in the spaces at the bottom of the page.
1
2
3
4
5
6
7
8
9 at a point 20 meters from a
flagpole, the angle of
10 if a rocket flies 2° off course for
1000 miles, how far from the
elevation of the top of
the flagpole is 48°.
correct path will the rocket be?
how tall is the
flagpole?

Explanation:

Step1: Recall the sine function

In a right - triangle, the sine of an angle $\theta$ is defined as $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. For problem (1), $\theta = 32^{\circ}$, the hypotenuse is $4$ m, and the side opposite to the $32^{\circ}$ angle is $x$. So, $\sin32^{\circ}=\frac{x}{4}$.

Step2: Solve for $x$

We know that $\sin32^{\circ}\approx0.5299$. Then $x = 4\times\sin32^{\circ}$. Substituting the value of $\sin32^{\circ}$, we get $x=4\times0.5299 = 2.1196\approx2.1$ m.

Step3: Recall the cosine function for problem (2)

In a right - triangle, the cosine of an angle $\theta$ is defined as $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$. For problem (2), $\theta = 57^{\circ}$, the hypotenuse is $22$ cm, and the side adjacent to the $57^{\circ}$ angle is $x$. So, $\cos57^{\circ}=\frac{x}{22}$.

Step4: Solve for $x$

We know that $\cos57^{\circ}\approx0.5446$. Then $x = 22\times\cos57^{\circ}$. Substituting the value of $\cos57^{\circ}$, we get $x=22\times0.5446=11.9812\approx12.0$ cm.

Step5: Recall the sine function for problem (3)

In a right - triangle, the sine of an angle $\theta$ is defined as $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. For problem (3), $\theta = 20^{\circ}$, the hypotenuse is $30$ m. Let the opposite side be $x$. So, $\sin20^{\circ}=\frac{x}{30}$.

Step6: Solve for $x$

We know that $\sin20^{\circ}\approx0.3420$. Then $x = 30\times\sin20^{\circ}$. Substituting the value of $\sin20^{\circ}$, we get $x=30\times0.3420 = 10.26\approx10.3$ m.

Step7: Recall the sine function for problem (5)

In a right - triangle, the sine of an angle $\theta$ is defined as $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. For problem (5), $\theta = 61^{\circ}$, the hypotenuse is $15$ cm. Let the opposite side be $x$. So, $\sin61^{\circ}=\frac{x}{15}$.

Step8: Solve for $x$

We know that $\sin61^{\circ}\approx0.8746$. Then $x = 15\times\sin61^{\circ}$. Substituting the value of $\sin61^{\circ}$, we get $x=15\times0.8746=13.119\approx13.1$ cm.

Step9: Recall the sine function for problem (6)

In a right - triangle, the sine of an angle $\theta$ is defined as $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. For problem (6), $\theta = 18^{\circ}$, the hypotenuse is $200$ m. Let the opposite side be $x$. So, $\sin18^{\circ}=\frac{x}{200}$.

Step10: Solve for $x$

We know that $\sin18^{\circ}\approx0.3090$. Then $x = 200\times\sin18^{\circ}$. Substituting the value of $\sin18^{\circ}$, we get $x=200\times0.3090 = 61.8$ m.

Step11: Recall the sine function for problem (7)

In a right - triangle, the sine of an angle $\theta$ is defined as $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. For problem (7), $\theta = 43^{\circ}$, the hypotenuse is $5$ m. Let the opposite side be $x$. So, $\sin43^{\circ}=\frac{x}{5}$.

Step12: Solve for $x$

We know that $\sin43^{\circ}\approx0.6820$. Then $x = 5\times\sin43^{\circ}$. Substituting the value of $\sin43^{\circ}$, we get $x=5\times0.6820=3.41\approx3.4$ m.

Step13: Recall the tangent function for problem (9)

In a right - triangle, the tangent of an angle $\theta$ is defined as $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. For problem (9), $\theta = 48^{\circ}$, the adjacent side is $20$ m. Let the opposite side (height of the flag - pole) be $x$. So, $\tan48^{\circ}=\frac{x}{20}$.

Step14: Solve for $x$

We know that $\tan48^{\circ}\approx1.1106$. Then $x = 20\times\tan48^{\circ}$. Substituting the value of $\tan48^{\circ}$, we get $x=20\times1.1106 = 22.212\approx22.2$ m.

Step15: Recall the sine function for problem (10)

In a right - triangle, the sine of an angle $\theta$ is defined as $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. For problem (10), $\theta = 2^{\circ}$, the hypotenuse is $1000$ mi. Let the opposite side (distance from the correct path) be $x$. So, $\sin2^{\circ}=\frac{x}{1000}$.

Step16: Solve for $x$

We know that $\sin2^{\circ}\approx0.0349$. Then $x = 1000\times\sin2^{\circ}$. Substituting the value of $\sin2^{\circ}$, we get $x=1000\times0.0349 = 34.9$ mi.

Answer:

1. $2.1$ m; 2. $12.0$ cm; 3. $10.3$ m; 5. $13.1$ cm; 6. $61.8$ m; 7. $3.4$ m; 9. $22.2$ m; 10. $34.9$ mi