Question

a triangular portion of a baseball field is marked as shown below. to the nearest tenth, what is the length of the side labeled ( c )?

diagram: triangle with vertices ( a ), ( b ), ( c ); angle at ( a = 36^circ ), angle at ( c = 28^circ ), side ( bc = 2 ) yds, side ( c ) ( ( ab ) ), side ( b ) ( ( ac ) ). note: \triangle not drawn to scale\.

options:

• a. ( 2.2 ) yds
• b. ( 3.4 ) yds
• c. ( 1.6 ) yds
• d. ( 2.9 ) yds

Explanation:

Step1: Find angle at B

The sum of angles in a triangle is \(180^\circ\). So, \(\angle B = 180^\circ - 36^\circ - 28^\circ = 116^\circ\).

Step2: Apply Law of Sines

Law of Sines: \(\frac{c}{\sin C}=\frac{BC}{\sin A}\). Here, \(BC = 2\) yds, \(\angle A = 36^\circ\), \(\angle C = 28^\circ\). So, \(\frac{c}{\sin 28^\circ}=\frac{2}{\sin 36^\circ}\).

Step3: Solve for c

\(c=\frac{2\times\sin 28^\circ}{\sin 36^\circ}\). Calculate \(\sin 28^\circ\approx0.4695\), \(\sin 36^\circ\approx0.5878\). Then \(c=\frac{2\times0.4695}{0.5878}\approx\frac{0.939}{0.5878}\approx1.6\) (Wait, no, wait, correction: Wait, angle C is 28, angle A is 36, side opposite angle A is BC? Wait no, in triangle ABC, side a is BC, side b is AC, side c is AB. Wait, Law of Sines: \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). So side opposite angle A (36°) is BC (length 2? Wait no, angle at A is 36°, angle at C is 28°, so side opposite angle A is BC, side opposite angle C is AB (which is c). So correct formula: \(\frac{c}{\sin C}=\frac{BC}{\sin A}\). So \(c = \frac{BC\times\sin C}{\sin A}\). Wait, BC is length 2, angle C is 28°, angle A is 36°. So \(c=\frac{2\times\sin 28^\circ}{\sin 36^\circ}\). Let's recalculate: \(\sin 28^\circ\approx0.4695\), \(\sin 36^\circ\approx0.5878\). So \(2\times0.4695 = 0.939\), divided by 0.5878: \(0.939\div0.5878\approx1.6\)? Wait but option C is 1.6, but wait maybe I mixed up angles. Wait, angle at B is 116°, side opposite angle B is AC (b), side opposite angle A (36°) is BC (2 yds), side opposite angle C (28°) is AB (c). Wait, maybe I made a mistake in angle assignment. Wait, let's re-express: In triangle ABC, vertices A, B, C. Angle at A: 36°, angle at C: 28°, so angle at B: 180 - 36 -28 = 116°. Side BC is opposite angle A (36°), so length BC = a = 2. Side AB is opposite angle C (28°), so length AB = c. So Law of Sines: \(\frac{a}{\sin A}=\frac{c}{\sin C}\). So \(c = \frac{a\times\sin C}{\sin A}=\frac{2\times\sin 28^\circ}{\sin 36^\circ}\). Calculating: \(\sin 28^\circ\approx0.4695\), \(\sin 36^\circ\approx0.5878\). So \(2\times0.4695 = 0.939\), \(0.939\div0.5878\approx1.6\). So c ≈1.6 yds.

Answer:

C. 1.6 yds