Question 766535: A surveyor stands at point A, which is due south of a tower OT of height h m. The angle of elevation of the top of the tower from A is 45 degrees. The surveyor then walks 100 m due east to point B, from where she measures the angle of elevation of the top of the tower to be 30 degrees.
P.S the answer says that the bearing is 130 degrees.
PLEASE PLEAS HELP ME I am really struggling to figure out how...
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! bearing is from where to where? ok, bearing required is from base of water tower to point B
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First consider the triangle made by points A, B and T (the top of the water tower) with A=45 degrees, B=30 degrees and T=105 degrees.
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Next, apply the law of sines to calculate b(distance of T to point A), also note that the distance from A to B is 100 meters.
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b/sin(30) = 100/sin(105)
b = 100*sin(30) / sin(105) = 51.76
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now consider the flat triangle AOB where O is the base of the water tower. We know that the length of AO is the height of the water tower because triangle AOT is an isosceles right triangle with two 45 degree angles and one 90 degree angle and b is its hypotenuse, so
51.76^2 = 2h^2
2679.1 = 2h^2
1339.5 = h^2
h = sqrt(1339.5) = 36.6 meters
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in right triangle AOB we have
tan(O) = 100 / 36.6 = 2.7 and angle O is 70 degrees
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due south bearing is 180 degrees so 180 - 70 = 110 degree bearing to B
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