# Precalculus

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1. A man stands 12 feet from a statue. The angle of elevation from the eye level to the top of the statue is 30 degrees, and the angle of depression to the base of the statue is 15. How tall in the statue?

2. Two boats lie on a straight line with the base of a lighthouse. From the top of the lighthouse, 21 meters above water level, it is observed that the angle of depression of the nearest boat is 53 degrees and the angle of depression of the farthers boat is 27 degrees. How far aprt are the boats?

• Precalculus - ,

#1 My diagram has two right angle triangles with a common right angle, the one with the 53 angle embedded within the larger one.

label the distance from the nearer boat to the lighthouse x, label the distance from the farther boat to the lighthouse y

then tan53 = 21/x
x = 21/tan53

and tan 27 = y/21

distance between the two boats = y-x

#1 (similar to #2)

my diagram has the man standing 12 to the left of the statue, A the top, B the bottom, the line to the top makes an angle of 30º and to the bottom of the statue 15º
draw a horizontalline from the "eye" to the statue meeting it at P
use the tangent ratio twice to find AP and BP
height of statue = AP + BP

• Precalculus - ,

i still am confused on what to do for the second explanation you gave

• Precalculus - ,

ok,
draw a triangle ABE with AB a vertical line, that is your statue.
E is to left of the line AB
Join EA and EB
From E draw a line perpendicular to AB to meet AB at P, P is between A and B
EP = 12

angle AEP=30º
angle BEP = 15º

in top triangle:
AP/12 = tan 30
AP = 12tan30 = 6.928

in bottom triangle
BP/12 = tan15
BP = 12tan15 = 3.215
AB = 10.14