Question

puma grove, the world’s best female golfer, is two shots away from winning another tournament.
from a marker exactly 200 meters from the pin, the angle between the pin and the ball is 110°, as shown.
from the ball, puma checks the angle between the marker and the pin, calculates, then swings. her ball travels 234 meters directly to the pin.
image of triangle with pin, marker (200m from pin), ball (234m from pin), angle 110° at marker between pin and ball
what was the measure of the angle between the marker and the pin?
do not round during your calculations. round your final answer to the nearest degree.

Explanation:

Step1: Identify the triangle sides and angle

We have a triangle with sides: \( a = 234 \) (Ball to Pin), \( b = 200 \) (Marker to Pin), \( c \) (Marker to Ball), and angle \( A = 110^\circ \) (at Marker, between Pin and Ball). We need to find angle \( B \) (at Ball, between Marker and Pin) using the Law of Sines: \( \frac{\sin A}{a}=\frac{\sin B}{b} \).

Step2: Apply Law of Sines

Substitute values: \( \frac{\sin 110^\circ}{234}=\frac{\sin B}{200} \). Solve for \( \sin B \): \( \sin B=\frac{200 \times \sin 110^\circ}{234} \). Calculate \( \sin 110^\circ \approx \sin(70^\circ) \approx 0.9396926 \). Then \( \sin B=\frac{200 \times 0.9396926}{234}=\frac{187.93852}{234}\approx 0.803156 \).

Step3: Find angle B

Take inverse sine: \( B = \arcsin(0.803156) \approx 53.4^\circ \), round to nearest degree: \( 53^\circ \). Wait, wait, no—wait, the angle at Ball: wait, let's recheck. Wait, the triangle: Marker (M), Ball (B), Pin (P). So \( \angle M = 110^\circ \), \( MP = 200 \), \( BP = 234 \), find \( \angle B \). Law of Sines: \( \frac{\sin \angle M}{BP}=\frac{\sin \angle B}{MP} \). So \( \sin \angle B=\frac{MP \times \sin \angle M}{BP}=\frac{200 \times \sin 110^\circ}{234} \). \( \sin 110^\circ \approx 0.9396926 \), so \( 200 \times 0.9396926 = 187.93852 \), divided by 234: \( 187.93852 / 234 \approx 0.803156 \). Then \( \angle B = \arcsin(0.803156) \approx 53.4^\circ \), but wait, is there a possibility of obtuse angle? Since \( \angle M = 110^\circ \) is obtuse, the other angles must be acute. So \( \angle B \approx 53^\circ \). Wait, but let's check again. Wait, maybe I mixed up the sides. Wait, \( MP = 200 \) (Marker to Pin), \( BP = 234 \) (Ball to Pin), \( MB \) is the side between Marker and Ball. So angle at M: \( \angle M = 110^\circ \), side opposite: \( BP = 234 \). Angle at B: \( \angle B \), side opposite: \( MP = 200 \). So Law of Sines: \( \frac{\sin \angle M}{BP}=\frac{\sin \angle B}{MP} \). So \( \sin \angle B = \frac{MP \times \sin \angle M}{BP} = \frac{200 \times \sin 110^\circ}{234} \). Calculating that: \( \sin 110^\circ \approx 0.9396926 \), so \( 200 * 0.9396926 = 187.93852 \), divided by 234: \( 187.93852 / 234 ≈ 0.803156 \). Then \( \angle B = \arcsin(0.803156) ≈ 53.4^\circ \), which rounds to 53 degrees. Wait, but let's confirm. Alternatively, maybe I made a mistake in the angle. Wait, the problem says "the angle between the marker and the pin" from the ball, so that's angle at Ball, between Marker and Pin, which is \( \angle B \). So yes, using Law of Sines, we get approximately 53 degrees. Wait, but let's check with calculator: \( \arcsin(0.803156) \). Let's compute 0.803156. \( \sin 53^\circ \approx 0.7986 \), \( \sin 54^\circ \approx 0.8090 \). So 0.803156 is between 53 and 54. Let's calculate the difference: 0.803156 - 0.7986 = 0.004556. The interval between 53 and 54 is 0.8090 - 0.7986 = 0.0104. So fraction: 0.004556 / 0.0104 ≈ 0.438. So approximately 53.44 degrees, which rounds to 53 degrees. Wait, but maybe I messed up the sides. Wait, the side from Ball to Pin is 234, Marker to Pin is 200, angle at Marker is 110 degrees. So triangle sides: MP = 200, BP = 234, angle at M is 110 degrees. So by Law of Sines, \( \frac{\sin \angle B}{MP} = \frac{\sin \angle M}{BP} \), so \( \sin \angle B = \frac{MP \times \sin \angle M}{BP} = \frac{200 \times \sin 110^\circ}{234} \). Let's compute \( \sin 110^\circ \): 110 degrees is in second quadrant, so \( \sin 110^\circ = \sin(70^\circ) ≈ 0.9396926208 \). Then 200 * 0.9396926208 = 187.93852416. Divide by 234: 187.93852416 / 234 ≈ 0.8031560862. Then \( \arcsin(0.8031560862) ≈ 53.436^\circ \), which rounds to 53 degrees. Wait, but maybe the problem is asking for the angle at the Ball, between Marker and Pin, which is what we calculated. So the answer is 53 degrees? Wait, no, wait—wait, maybe I mixed up the angle. Wait, the sum of angles in a triangle is 180. So angle at M is 110, angle at B is ~53, so angle at P is 180 - 110 - 53 = 17? That seems small, but maybe. Wait, let's check with Law of Sines for angle at P. \( \frac{\sin \angle P}{MB} = \frac{\sin \angle M}{BP} \). But we don't know MB. Alternatively, maybe I made a mistake in the side labels. Wait, the marker is 200 meters from the pin (MP = 200), the ball is 234 meters from the pin (BP = 234), and the angle at the marker between pin and ball is 110 degrees (angle at M: \( \angle PMB = 110^\circ \)). So triangle PMB: sides PM = 200, PB = 234, angle at M is 110 degrees. We need to find angle at B: \( \angle PBM \), which is the angle between marker (M) and pin (P) from the ball (B). So using Law of Sines: \( \frac{\sin \angle PBM}{PM} = \frac{\sin \angle PMB}{PB} \). So \( \sin \angle PBM = \frac{PM \times \sin \angle PMB}{PB} = \frac{200 \times \sin 110^\circ}{234} \), which we calculated as ~0.803, so angle ~53 degrees. So the answer is 53 degrees.

Answer:

\( 53 \)