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?

The altitude of the climber is 1.40 miles

Step-by-step 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

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.

6 spoons of sugar for every 21 cups of flour

step-by-step explanation: