THIS is about TRIGONOMETRY pls do answer......trigo functions and problem solving?

Question

Find the reference angle of each angle theta, and then find the exact values of the six trigonometric funtions of theta.

1.) theta = 150 degrees
2.) theta = 9pi/4
3.) theta = -2pi/3

PROBLEM SOLVING

1.) A commercial jet is flying at a constant speed on a level course that will carry it directly over Albuquerque, New mexico. Through a hole in a cloud cover, the captain obsrves the light of the city at the angle depression of 7-degree. 3 minutes later, the captain catches a second glimpse, now an angle of depression 13-degree. In how many more minutes will the jet be directly over the city?

2.)Two coast guards station A and B are located along a straight coastline. Station A is 20 miles due to north of Station B. A supertanker in distress is obsevred from station A at a bearing of S 9degree-10' E, from station B at a bearing of S 37degrees-40' E. Find the distance from the supertanker to the nearest point of the coastline...

Pls answer wat u can answer...really need your help

Answer 1

The "reference angle" is the smallest positive angle that can be made with the given angle and either the positive or negative x-axis. So the reference angles are:

(1) 150 -> 30 degrees (above negative x-axis)
(2) 9pi/4 -> pi/4 radians (above positive x-axis)
(3) -2pi/3 -> pi/3 radians (below negative x-axis)

30 degrees: sin = 1/2, cos = sqrt(3)/2, tan = 1/sqrt(3), sec = 2/sqrt(3), csc = 2, cot = sqrt(3)

pi/4 radians = 45 degrees: sin = sqrt(2)/2, cos = sqrt(2)/2, tan = 1, sec = sqrt(2), csc = sqrt(2), cot = 1

pi/3 radians = 60 degrees: sin = sqrt(3)/2, cos = 1/2, tan = sqrt(3), sec = 2, csc = 2/sqrt(3), cot = 1/sqrt(3)

By the way, it's a good idea to commit those values to memory. 0 - 30 - 45 - 60 - 90 degrees have sine:
sqrt(0)/2 [= 0],
sqrt(1)/2 [= 1/2],
sqrt(2)/2,
sqrt(3)/2, and
sqrt(4)/2 [= 1], respectively.

And they have the same sequence for cosine, but in the opposite order: 1, sqrt(3)/2, sqrt(2)/2, 1/2, and 0 respectively. From those two lists you can construct all the rest.

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Problems:

(1) Given an altitude (h), and an angle of depression (x), and distance to the city along the ground (d), you have a right triangle where:

tan(x) = h/d

when x=7 degrees, d = 8.14h
when x=13 degrees, d = 4.33h

So, in 3 minutes the plane covers (8.14-4.33)h of the distance, which is 3.81h. If the plane covers 3.81h in 3 minutes, it should take:

t / (4.33h) = 3 / (3.81h)
t = 3 * 4.33 / 3.81
t = 3.41 minutes

... to complete the remaining 4.33h.

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(2) There are two right triangles, involving the shore, the distance to the ship, and the lines of sight to the ship.

The perpendicular distance to the ship strikes the shoreline some distance south of station B, Let's call the distance on the shore (from B) s, and the distance the ship is out to sea, d.

B's right triangle:

tan(37.6667) = d / s

A's right triangle:

tan(9.1667) = d / (s + 20)

Now you have two equations in two variables and just need to solve them:

tan(37.6667) = d / s
0.771959 = d / s
d = 0.771959 s

tan(9.1667) = d / (s + 20)
0.161368 = d / (s + 20)
d = 0.161368 s + 3.22735

Since both equal d, let's solve for s:

0.771959 s = 0.161368 s + 3.22735
0.610591 s = 3.22735
s = 5.28562

Use that value for s to solve for d:

d = 0.771959 s
d = 0.771959 * 5.28562
d = 4.08 miles