Answer to Question #177551 in Trigonometry for Sree

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

"CB" is the distance between the ball and the hole.


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

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


2) Using Pythagorean's theorem "a^2+b^2=\u0441^2".

Then "CH^2 + AH^2 = CA^2"

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

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


3) "BH = BA - AH"

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


4) Using Pythagorean's theorem "a^2+b^2=\u0441^2".

Then "CH^2 + BH^2 = CB^2"

"CB = \\sqrt {CH^2 + BH^2}"

"CB = \\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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