Marc P. answered • 11/17/19

Ivy League Grad to Help with Standardized Tests and Academic Work

This problem is a little more complicated than the other one you posted. Let's call the location of the first team point A, the location of the second team point B, and the location of the climber point C. We can draw a triangle with points A and B along the bottom. Because we know that their altitude is 1 mile, in order to find the altitude of the climber we need the height of the triangle. We have the base of the triangle (side c), which is .5, so let's try to get side b. First we need to get angle C. Because we know the other two angles, we find C = 180 - 14 - 22 = 144. Now we can use the law of sines, which says b/sinB = c/sinC. We can rearrange this to get b = c*sinB/sinC. Because we know the length of side c and the measures of angles B and C, we can get the length of side b. Now draw the triangle's height, a vertical line from point C down to a point on the base (let's call it point D). Now use the law of sines on triangle ACD:

a'/sinA = d/sinD [I'm calling the height a', since it is the side opposite angle A in the small triangle]

a' = d*sinA/sinD [multiplying both sides by sinA]

a' = b*sinA/sinD [since side b is the same as side d]

a' = b*sinA [since D = 90, so sinD = 1]

You know b and A, so now you can calculate a'.