At a point 30 m from a building, the angle of elevation to the top of the building is 70 degrees. If the observer's eyes is 1.2m above ground, how tall is the building?
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First try drawing a diagram. The base is 30m and the angles we know are 90 degrees at the base of the building and 70 degrees away from the building. That leaves 180 - 90 - 70 = 20 degrees at the top. Ignore the 1.2m for now; we will add it in later.
The tangent trigonometry function can give the answer.
tan = opp / adj
tan( 70 degrees ) = height / 30m
30 m tan(70 degrees) = height
height = 82.42 m
Add the 1.2 m of the observer's eye to get: 82.42 m + 1.2 m = 83.6 m
The tangent trigonometry function can give the answer.
tan = opp / adj
tan( 70 degrees ) = height / 30m
30 m tan(70 degrees) = height
height = 82.42 m
Add the 1.2 m of the observer's eye to get: 82.42 m + 1.2 m = 83.6 m
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30tan70 + 1.2?
i did simple trigonometry using sohcahtoa.....
do a triangle and have it 1.2m above ground level. make the unknown side x and add 1.2 after.....
im not sure if its right.....check it with someone....
i did simple trigonometry using sohcahtoa.....
do a triangle and have it 1.2m above ground level. make the unknown side x and add 1.2 after.....
im not sure if its right.....check it with someone....
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tan 70 ° = x / 30
x = 30 tan 70 °
x = 82.4 m
Height of building = 82.4 + 1.2 m = 83.6 m
x = 30 tan 70 °
x = 82.4 m
Height of building = 82.4 + 1.2 m = 83.6 m