1st we draw a picture of the prob.
1. Draw a ver and hor line to form a rt
angle.
2. Draw the hyp. from top of ver line to the end of hor line.
The angle between the hyp and hor line = 54 deg.
3. Draw a 2nd hyp from the top of ver
line to a point on the hor line.The
distance from this point to bottom of
the ver line is 270m. The angle formed
is 71 deg. The dist. between the 2 points where the 2 hyp cross the hor
line is X.
tan54 = h / (270+X),
h = (270 + X)tan54,
tan71 = h / 270,
h = 270tan71,
h = (270 + X)tan54 = 270tan71,
Divide both sides by tan54:
270 + X = 270tan71 / tan54 = 569.71,
X = 569.71 - 270 = 299.7m.
An observer in a hot air balloon some distance away from the aqueduct determines that the angle of each depression to each end is 54 degrees and 71 degrees. The closest end of the aqueduct is 270 m from the balloon. Calculate the length of the aqueduct to the nearest tenth of a metre
Answer: 299.8 m
I confused, i tried using sine law
these were my calculations but some how i'm not getting the right answer
sin54/270=sin 109/x
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