In a tournament, a professional golfer knows that she is 200 yards from the hole. A spectator is watching her play and is 140 yards away from the golfer. If the spectator has an angle of 110° between the golfer and the hole, what is the angle that the golfer has between the spectator and the hole? 70.0° 41.1° 28.9° 19.9°

The golfer, hole and spectator form a triangle. Let ABC be the triangle and let the angle the spectator has between the golfer and the hole be A = 110°, the angle the golfer has between the spectator and the hole be B and the angle the hole has between the golfer and the spectator be C. Let the angle between the golfer and the hole be a = 200 yards, the distance between the spectator and the hole be b and the distance between the golfer and the spectator be c = 140 yards,

Using the sine rule for the triangle, we find angle C.

So, a/sinA = b/SinB = c/SinC

So, a/sinA = c/sinC

sinC = csinA/a

C = sin⁻¹(csinA/a)

Substituting the values of the variables into the equation, we have

C = sin⁻¹(csinA/a)

C = sin⁻¹( 140sin110°/200)

C = sin⁻¹( 7 × 0.9397/10)

C = sin⁻¹(6.5778/10)

C = sin⁻¹(0.65778)

C = 41.13°

We know that A + B + C = 180° (sum of angles in a triangle)

And since the angle the golfer has between the spectator and the hole be B

So, B = 180° - (A + C)

B = 180° - (110° + 41.13°)

B = 180° - 151.13°

B = 28.87°

B ≅ 28.9°

abiolataiwo2015 abiolataiwo2015 · Answer 2 · 2022-01-14T12:22:13+01:00

The angle that the golfer has between the spectator and the hole is 28.9°.

Angel between spectator and holes:

Using the law of sines:

BC/SinΔBAC=AB/SinΔACB

Hence,

200/Sin110°=140/SinΔACB

ΔACB=41.1°

Thus,

ΔABC=180°-ΔBAC-ΔACB

ΔABC= 180° - (110° + 41.1°)

ΔABC= 180° - 151.1°

ΔABC= 28.9°

Inconclusion the angle that the golfer has between the spectator and the hole is 28.9°.

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