Asked by Evans
It was observed that the shadow of a vertical pole of 6 meters longer when the angle of elevation of the sun was 30° than when it was 60°.By means of a sketched diagram,calculate,correct to two decimal places,the height of the pole.
Answers
Answered by
John
tan 60deg = y/x y/x =1.7325
tan 30 deg = y/(x + 6) = .577
x = shadow and y = pole
y = 1.7325x
y = .577(x +6)
Since they both equal y, you can say..
1.7325x = .577x + 3.462
Solve for x to get the shadow for the 60 degrees.
Find y, by putting this answer into both equations. If you did everything correct, you should get the same value for y. I got around 5.2 meters. See if you get the same thing.
tan 30 deg = y/(x + 6) = .577
x = shadow and y = pole
y = 1.7325x
y = .577(x +6)
Since they both equal y, you can say..
1.7325x = .577x + 3.462
Solve for x to get the shadow for the 60 degrees.
Find y, by putting this answer into both equations. If you did everything correct, you should get the same value for y. I got around 5.2 meters. See if you get the same thing.
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