Asked by Anonymous
When the angle of elevation of the sun is 30° the shadow of a vertical tower is 20m longer than when the elevation of the sun is 60°.find the height of the tower
Answers
Answered by
Reiny
After making your sketch, .....
Let the shadow's length at 60° be x m, let the tower height be h
Given: the length of the shadow at 30° = x+20 m
so ... tan60 = h/x
h = xtan60
and tan 30 = h/(x+20)
h = tan30(x+20)
then xtan60 = tan30(x+20)
xtan60 = xtan30 + 20tan30
x(tan60 - tan30) = 20tan30
x = 20tan30/((tan60 - tan30) = ....
then you find h in h = xtan60
OR
Nice to have those 30° and 60° angles, since you have the 30-60-90 triangle
with corresponding sides in the ratio of 1:√3:2
By the 30-60-90 ratios:
√3 x / x = (x+2)/(√3x)
√3 = (x+20)/(√3x)
3x = x+20
x = 10
then h = √3x = √3(10) = appr 17.321.... m
You will get the same answer from my first solution.
Let the shadow's length at 60° be x m, let the tower height be h
Given: the length of the shadow at 30° = x+20 m
so ... tan60 = h/x
h = xtan60
and tan 30 = h/(x+20)
h = tan30(x+20)
then xtan60 = tan30(x+20)
xtan60 = xtan30 + 20tan30
x(tan60 - tan30) = 20tan30
x = 20tan30/((tan60 - tan30) = ....
then you find h in h = xtan60
OR
Nice to have those 30° and 60° angles, since you have the 30-60-90 triangle
with corresponding sides in the ratio of 1:√3:2
By the 30-60-90 ratios:
√3 x / x = (x+2)/(√3x)
√3 = (x+20)/(√3x)
3x = x+20
x = 10
then h = √3x = √3(10) = appr 17.321.... m
You will get the same answer from my first solution.
Answered by
oobleck
Reiny's first solution can be made at least to look less complicated if you are comfortable using the cotangent function.
h cot30° - h cot60° = 20
h = 20/(cot30° - cot60°) = 20/(√3 - 1/√3)
h cot30° - h cot60° = 20
h = 20/(cot30° - cot60°) = 20/(√3 - 1/√3)
Answered by
Ayoola joy
I need the real solution to the question
Answered by
Esther
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