WHAT IS THE ALTITUDE OF THE CLIMBER

Question

Two search teams spot a stranded climber on a mountain. The first search team is 0.5 miles from the second search team, and both teams are on the same side of the mountain at an altitude of 1 mile. The angle of elevation from the first search team to the stranded climber is 14°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile.

Answer 1

draw a diagram

triangle ABC with the climber at C

the altitude of the climber is CX where X is the point collinear with points A and B

angle CAB=14 degrees

angle CBX=22 degrees

tan=opposite/adjacent

call BX length "h"

tan14°=h/(0.5+x)

tan22°=h/x

0.249=h/(0.5+x)

0.404=h/x

0.249(0.5+x)=h

0.404x=h

0.404x=0.249(0.5+x)

0.404x=0.1245+0.249x

0.155x=0.1245

155x=124.5

x=124.5/155

x=0.803 miles

AX=0.5+0.803

AX=1.303

sin14degrees=h/1.303

0.249=h/1.303

0.249*1.303=h

h=0.324 miles above the search teams

the search teams are 1 mile up the mountain

the altitude of the stranded climber is 1.324 miles

Answer 2

I guess the teams and the climber are in a straight line.

a=.5

γ=166°

α=22°

a/sinα=1.3347.....=b/sinβ=c/sinγ

=c/.2419218......=1.33*166=322.89=c

a^2+b^2=c^2

322.89^2-22^2=b^2 so 322.14=b

Now we need a rt triangle so side b is now side c of it.

The new side b is b/sinβ=322.14

sin of a rt angle=1

b=322.14*sin14º(now it'sβ)=77.93miles