Question

directions: give each trigonometric ratio as a fraction in simplest form.
1.

• sin a =
• cos a =
• tan a =
• sin c
• cos c
• tan c

2.

• sin w =
• cos w =
• tan w =
• sin x =
• cos x =
• tan x =

3.

• sin l =
• cos l =
• tan l =
• sin m =
• cos m =
• tan m =

Explanation:

Step1: Recall trigonometric ratios

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\)

For problem 1:

• Hypotenuse \(AC = 13\), \(AB = 12\), \(BC = 5\)
• For \(\angle A\):
• Opposite side to \(\angle A\) is \(BC = 5\), adjacent side to \(\angle A\) is \(AB=12\)
• \(\sin A=\frac{BC}{AC}=\frac{5}{13}\)
• \(\cos A=\frac{AB}{AC}=\frac{12}{13}\)
• \(\tan A=\frac{BC}{AB}=\frac{5}{12}\)
• For \(\angle C\):
• Opposite side to \(\angle C\) is \(AB = 12\), adjacent side to \(\angle C\) is \(BC = 5\)
• \(\sin C=\frac{AB}{AC}=\frac{12}{13}\)
• \(\cos C=\frac{BC}{AC}=\frac{5}{13}\)
• \(\tan C=\frac{AB}{BC}=\frac{12}{5}\)

For problem 2:

• First, find the hypotenuse \(WX\) using Pythagorean theorem \(WX=\sqrt{9^{2}+12^{2}}=\sqrt{81 + 144}=\sqrt{225}=15\)
• For \(\angle W\):
• Opposite side to \(\angle W\) is \(XY = 12\), adjacent side to \(\angle W\) is \(WY = 9\)
• \(\sin W=\frac{XY}{WX}=\frac{12}{15}=\frac{4}{5}\)
• \(\cos W=\frac{WY}{WX}=\frac{9}{15}=\frac{3}{5}\)
• \(\tan W=\frac{XY}{WY}=\frac{12}{9}=\frac{4}{3}\)
• For \(\angle X\):
• Opposite side to \(\angle X\) is \(WY = 9\), adjacent side to \(\angle X\) is \(XY = 12\)
• \(\sin X=\frac{WY}{WX}=\frac{9}{15}=\frac{3}{5}\)
• \(\cos X=\frac{XY}{WX}=\frac{12}{15}=\frac{4}{5}\)
• \(\tan X=\frac{WY}{XY}=\frac{9}{12}=\frac{3}{4}\)

For problem 3:

• First, find the third side \(KM\) using Pythagorean theorem \(KM=\sqrt{34^{2}-16^{2}}=\sqrt{(34 + 16)(34 - 16)}=\sqrt{50\times18}=\sqrt{900}=30\)
• For \(\angle L\):
• Opposite side to \(\angle L\) is \(KM = 30\), adjacent side to \(\angle L\) is \(KL = 16\)
• \(\sin L=\frac{KM}{LM}=\frac{30}{34}=\frac{15}{17}\)
• \(\cos L=\frac{KL}{LM}=\frac{16}{34}=\frac{8}{17}\)
• \(\tan L=\frac{KM}{KL}=\frac{30}{16}=\frac{15}{8}\)
• For \(\angle M\):
• Opposite side to \(\angle M\) is \(KL = 16\), adjacent side to \(\angle M\) is \(KM = 30\)
• \(\sin M=\frac{KL}{LM}=\frac{16}{34}=\frac{8}{17}\)
• \(\cos M=\frac{KM}{LM}=\frac{30}{34}=\frac{15}{17}\)
• \(\tan M=\frac{KL}{KM}=\frac{16}{30}=\frac{8}{15}\)

Answer:

1.

• \(\sin A=\frac{5}{13}\), \(\cos A=\frac{12}{13}\), \(\tan A=\frac{5}{12}\)
• \(\sin C=\frac{12}{13}\), \(\cos C=\frac{5}{13}\), \(\tan C=\frac{12}{5}\)

2.

• \(\sin W=\frac{4}{5}\), \(\cos W=\frac{3}{5}\), \(\tan W=\frac{4}{3}\)
• \(\sin X=\frac{3}{5}\), \(\cos X=\frac{4}{5}\), \(\tan X=\frac{3}{4}\)

3.

• \(\sin L=\frac{15}{17}\), \(\cos L=\frac{8}{17}\), \(\tan L=\frac{15}{8}\)
• \(\sin M=\frac{8}{17}\), \(\cos M=\frac{15}{17}\), \(\tan M=\frac{8}{15}\)