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A word problem in geometry of mathmatics :[
Can't seem to figure this out, I just don't understand how to draw out the diagram.
From a point a certain distance away from a church, the angle of elevation to the church steeple
is 20 degrees. From a point 40m further away, the angle of elevation to the steeple is 33 degrees.
Determine the height of the steeple.
Solution
Let the A be the point on the ground, G be the base of the church and S be the church steeple.
So AGH is right angled triangle with angle AGS = 90 degrees.
So AG*tan GAS = GS.
=> AG tan 20 = GS.
=> AG = GS/tan20.....(1)
When we move by 40 meter further to a point B along the line AG , The angle GBS = 33 deg.
So BG tan GBP = GS.
BG = GS/tan3....(2).
AB- BG = GS(1/tan 20 - 1/tan33).
AB-BG = 40 m = GS(1/tan20-tan33).
Therefore GS = 40 /(1/tan20-1/tan33) = 33.12 meter.
Therefore the height of the tower = GS = 32.12 m.

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A word problem in geometry of mathmatics [Cant seem to figure t.pdf

  • 1. A word problem in geometry of mathmatics :[ Can't seem to figure this out, I just don't understand how to draw out the diagram. From a point a certain distance away from a church, the angle of elevation to the church steeple is 20 degrees. From a point 40m further away, the angle of elevation to the steeple is 33 degrees. Determine the height of the steeple. Solution Let the A be the point on the ground, G be the base of the church and S be the church steeple. So AGH is right angled triangle with angle AGS = 90 degrees. So AG*tan GAS = GS. => AG tan 20 = GS. => AG = GS/tan20.....(1) When we move by 40 meter further to a point B along the line AG , The angle GBS = 33 deg. So BG tan GBP = GS. BG = GS/tan3....(2). AB- BG = GS(1/tan 20 - 1/tan33). AB-BG = 40 m = GS(1/tan20-tan33). Therefore GS = 40 /(1/tan20-1/tan33) = 33.12 meter. Therefore the height of the tower = GS = 32.12 m.