Question

a golfer hits a tee shot on a 285 m long straight golf hole. the ball is hooked (hit at an angle) 14 degrees to the left. the ball lands 183 m from the tee. how far is the ball from the hole? show a diagram as well

Answer 1

Answer: about 116.2 m from the hole.

Work (two ways):

1) Law of cosines
- Triangle sides: Tee→Hole = 285 m, Tee→Ball = 183 m, included angle = 14°.
- Distance Ball→Hole = d where
d^2 = 285^2 + 183^2 − 2·285·183·cos(14°).
- Compute:
285^2 = 81,225
183^2 = 33,489
2·285·183·cos(14°) ≈ 104,310·0.970296 ≈ 101,211.55
d^2 ≈ 81,225 + 33,489 − 101,211.55 = 13,502.45
d ≈ sqrt(13,502.45) ≈ 116.19 m

So the ball is about 116.2 m from the hole.

2) Coordinate method (gives same result)
- Put tee at (0,0) and hole at (285,0). Ball at (183 cos14°, 183 sin14°) ≈ (177.56, 44.27).
- Vector Hole − Ball = (285 − 177.56, 0 − 44.27) ≈ (107.44, −44.27).
- Distance = sqrt(107.44^2 + 44.27^2) ≈ 116.19 m.

Simple diagram (not to scale):

T (tee) at origin ----→ H (hole, 285 m)
\
\
B (ball, 183 m at 14° left)

Or with coordinates:

T(0,0)
\
\
B(177.56, 44.27)
\
\
H(285, 0)

(Answer ≈ 116.2 m.)