trig probs: dist. from observer to plane; of boat from tower

[Mole]

Express your answer with the correct units and number of significant figures.

1. An airplane is flying at an altitude of 8000 m above the ground. An observer on the ground, whose eyes are 1.60 m above the ground, measures the angle to plane at several points. Calculate the straight-line distance from the observer to the plane when the angle is a) 30 degrees b) 75, and c) 90

I know the answers, but I don't understand how 90 degrees works with sin=opposite/hypotenuse, cos=adjacent/hypotenuse, tan=opposite/adjacent. The opposite side is the hypotenuse, so opposite/hypotenuse = 1, but what about adjacent and opposite? Which side is opposite and which side is adjacent?

5. A Coast Guard officer on the top of a 50 m tower sees a boat in the ocean. The boat is at an angle of 4 degrees below the horizontal. How far away is the boat from the base of the tower?

English is not my first language. Is horizontal the ocean horizon? I'm supposed to ignore the curve if I'm only using trig, so what angle is formed with the man and the boat? 64 degrees, 56 degrees, 34 degrees, 26 degrees?

 

[mmm4444bot]

Triangles, and a line ...

Hello Mole:

Opposite and adjacent work in situations where you have a triangle. When the airplane is directly above the observer, there is no triangle; the straight-line distance is just an 7,998.4-meter line.

The horizontal refers to the Coast Guard officer's line-of-sight parallel to the ocean's surface.

~ Mark

My edits: Corrected value, and then corrected the units (good grief)

 

[soroban]

Re: Two trig problems

Hello, Mole!

Did you make a sketch?

1. An airplane is flying at an altitude of 8000 m above the ground.
An observer on the ground, whose eyes are 1.60 m above the ground,
measures the angle to plane at several points.

Calculate the straight-line distance from the observer to the plane when the angle is:
. . \(\displaystyle a)\;30^o\qquad b)\;75^o\qquad c)\;90^o\)

      P *
        |   *
        |       *    d
 7998.4  |           *
        |               *
        |                   *
      B * - - - - - - - - - - - * A

\(\displaystyle \text{The plane is at }P\text{, the observer is }A.\)

\(\displaystyle \text{The plane is 7998.4 m above his eyes.}\)

\(\displaystyle \text{Let }d\text{ = distance }AP.\)

\(\displaystyle \text{In right triangle }PBA\!:\;\;\sin A \:=\:\frac{7998.4}{d} \quad\Rightarrow\quad d \:=\:\frac{7998.4}{\sin A}\)

\(\displaystyle \text{Use this formula for (a), (b), (c).}\)