Answer to Question #177551 in Trigonometry for Sree

Question

 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?

Answer 1

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 $a^2+b^2=с^2$.

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 $a^2+b^2=с^2$.

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.