Posted by Mz. Suzie on Monday, March 2, 2009 at 2:30pm.
#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
i still am confused on what to do for the second explanation you gave
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
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