Question

evaluate in fraction form
sin(a)=
cos(a)=
tan(a)=
sin(b)=
cos(b)=
tan(b)
(right triangle with legs 5, 12 and hypotenuse 13, vertices a, b, c with right angle at c)

Explanation:

Step1: Recall Trigonometric Ratios

In a right triangle, for an acute angle:

• $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$
• $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$
• $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$

Given right triangle \( \triangle ABC \) with \( \angle C = 90^\circ \), \( AC = 5 \), \( BC = 12 \), \( AB = 13 \) (hypotenuse, since \( 5^2 + 12^2 = 13^2 \)).

For \( \angle A \):

• Opposite side to \( \angle A \): \( BC = 12 \)
• Adjacent side to \( \angle A \): \( AC = 5 \)
• Hypotenuse: \( AB = 13 \)

Step2: Calculate \( \sin(A) \)

\( \sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{12}{13} \)

Step3: Calculate \( \cos(A) \)

\( \cos(A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AC}{AB} = \frac{5}{13} \)

Step4: Calculate \( \tan(A) \)

\( \tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AC} = \frac{12}{5} \)

For \( \angle B \):

• Opposite side to \( \angle B \): \( AC = 5 \)
• Adjacent side to \( \angle B \): \( BC = 12 \)
• Hypotenuse: \( AB = 13 \)

Step5: Calculate \( \sin(B) \)

\( \sin(B) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AC}{AB} = \frac{5}{13} \)

Step6: Calculate \( \cos(B) \)

\( \cos(B) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{12}{13} \)

Step7: Calculate \( \tan(B) \)

\( \tan(B) = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{BC} = \frac{5}{12} \)

Answer:

• \( \sin(A) = \boldsymbol{\frac{12}{13}} \)
• \( \cos(A) = \boldsymbol{\frac{5}{13}} \)
• \( \tan(A) = \boldsymbol{\frac{12}{5}} \)
• \( \sin(B) = \boldsymbol{\frac{5}{13}} \)
• \( \cos(B) = \boldsymbol{\frac{12}{13}} \)
• \( \tan(B) = \boldsymbol{\frac{5}{12}} \)