Question

Two search teams spot a stranded climber on a mountain. The first search team is 0.5 miles from the second search team. If the angle of elevation from the first search team to the stranded climber is 15° and the angle of elevation from the second search team is 22° to the stranded climber, what is the altitude of the climber if both search teams are standing at an altitude of 1 mile high?

Answer 1

Explanation:

Just solved a similar problem and figured it out, the angle of elevation is somewhat unintuitive as the angle of the second search team needs to be flipped. The diagram should look more like this:
This isn't an explanation of the math but more of a visualization for those that just needed the 2D representation. The real answer would be 1.081 miles (:

Answer 2

The altitude of the climber is 1.40 miles

Explanation:

see the attached figure to better understand the problem

we know that

In the right triangle BCD

tan(22°)=h/(x-0.5)

h=tan(22°)*(x-0.5) ----> equation A

In the right triangle ACD

tan(15°)=h/x

h=tan(15°)*(x) ----> equation B

Equate equation A and equation B and solve for x

tan(22°)*(x-0.5)=tan(15°)*(x)

tan(22°)*x-tan(22°)*0.5=tan(15°)*x

x[tan(22°)-tan(15°)]=tan(22°)*0.5

x=tan(22°)*0.5/[tan(22°)-tan(15°)]

x=1.48 miles

Find the value oh h

h=tan(15°)*(1.48)=0.40 miles

therefore

The altitude of the climber is equal to

0.40+1=1.40 miles