Question #177551

 Golfer hits his ball B a distance of 170m towards a Hole H which measures 195m from the Tee T to the green. If his shot is directed 10 degrees away from the true line to the hole, find the distance between his ball and the hole?


1
Expert's answer
2021-04-15T07:52:00-0400

Let AHAH is the length from the Golfer to the perpendicular going off to the ball, CA=170CA = 170 m, BA=195BA = 195 m, A=10°\angle {A} = 10\degree.

CBCB is the distance between the ball and the hole.


1) Using right triangle trigonometry: cosx=adjacenthypotenuse\cos{x} = \frac {adjacent}{hypotenuse} and adjacent=hypotenusecosxadjacent = hypotenuse *\cos{x}.

Then AH=CAcosA=170cos10°=167,4AH = CA * \cos{\angle {A} } = 170 * \cos{10\degree } = 167,4 m.


2) Using Pythagorean's theorem a2+b2=с2a^2+b^2=с^2.

Then CH2+AH2=CA2CH^2 + AH^2 = CA^2

CH=CA2AH2CH = \sqrt {CA^2 - AH^2}

CH=1702167,42=29,5CH = \sqrt {170^2 - 167,4^2} = 29,5 m


3) BH=BAAHBH = BA - AH

BH=195167,4=27,6BH = 195 - 167,4 = 27,6 m


4) Using Pythagorean's theorem a2+b2=с2a^2+b^2=с^2.

Then CH2+BH2=CB2CH^2 + BH^2 = CB^2

CB=CH2+BH2CB = \sqrt {CH^2 + BH^2}

CB=29,52+27,62=40,4CB = \sqrt {29,5^2 + 27,6^2} = 40,4 m


Hence the distance between the ball and the hole is 40,4 m.




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