Question

1. a boy flying a kite lets out 100 feet of string making an angle of elevation of 40°. how high above the ground is the kite? 9. as viewed from a cliff 360 m above sea level, the angle of depression to a ship is 28°. how far is the ship from the shore? 10. the angle of elevation from a ship to the top of a lighthouse is 3°. if the ship is 1,000 km from the lighthouse, how tall is the lighthouse? 11. a person hang gliding at an altitude of 300 feet is over a spot 2,250 feet from an area of soft grass where he would like to land. at what angle of depression should he see the grass?* 12. a 12 ft guy - wire is attached to a telephone pole at a point 9.5 feet below the top of the pole. if the wire forms a 51° angle with the ground, how high is the telephone pole? 13. from a lighthouse 1,000 ft above sea level, the angle of depression to a boat is 29°. the boat is moving forward and one minute later the angle of depression is 44°. how far has the boat traveled?(work problem twice once with each angle and compare)

Explanation:

Step1: 문제 8번 풀이

사인 함수 사용. $\sin\theta=\frac{opposite}{hypotenuse}$로, $\theta = 40^{\circ}$, $hypotenuse = 100$피트 이므로, $opposite = 100\times\sin40^{\circ}$.
$100\times\sin40^{\circ}\approx100\times0.6428 = 64.28$피트.

Step2: 문제 9번 풀이

탄젠트 함수 사용. $\tan\theta=\frac{opposite}{adjacent}$로, $\theta = 28^{\circ}$, $opposite = 360$미터 이므로, $adjacent=\frac{360}{\tan28^{\circ}}$.
$\frac{360}{\tan28^{\circ}}\approx\frac{360}{0.5317}\approx677$미터.

Step3: 문제 10번 풀이

탄젠트 함수 사용. $\tan\theta=\frac{opposite}{adjacent}$로, $\theta = 3^{\circ}$, $adjacent = 1000$킬로미터 이므로, $opposite = 1000\times\tan3^{\circ}$.
$1000\times\tan3^{\circ}\approx1000\times0.0524 = 52.4$킬로미터.

Step4: 문제 11번 풀이

탄젠트 함수 사용. $\tan\theta=\frac{opposite}{adjacent}$로, $opposite = 300$피트, $adjacent = 2250$피트 이므로, $\theta=\arctan(\frac{300}{2250})$.
$\theta=\arctan(\frac{300}{2250})\approx\arctan(0.1333)\approx7.59^{\circ}$.

Step5: 문제 12번 풀이

탄젠트 함수로 전화주 전체 높이를 구함. $\tan51^{\circ}=\frac{h}{12}$에서 $h = 12\times\tan51^{\circ}$.
$h\approx12\times1.2349 = 14.82$피트. 전화주 높이는 $14.82 + 9.5=24.32$피트.

Step6: 문제 13번 풀이

먼저 $\theta = 29^{\circ}$일 때, $\tan29^{\circ}=\frac{1000}{d_1}$에서 $d_1=\frac{1000}{\tan29^{\circ}}\approx\frac{1000}{0.5543}\approx1804$피트.
$\theta = 44^{\circ}$일 때, $\tan44^{\circ}=\frac{1000}{d_2}$에서 $d_2=\frac{1000}{\tan44^{\circ}}\approx\frac{1000}{0.9657}\approx1036$피트.
배가 이동한 거리는 $d_1 - d_2\approx1804 - 1036 = 768$피트.

Answer:

1. 64.28피트
2. 677미터
3. 52.4킬로미터
4. $7.59^{\circ}$
5. 24.32피트
6. 768피트