Question

similar right triangles - introduction to trigonometry
skills
for problems 1 - 6, use the triangle to the right to find the given trigonometric ratios.

1. cos n=
2. sin n=
3. tan n=
4. sin p=
5. cos p=
6. tan p=
7. given the right triangle shown, which of the following represents the value of tan a?

(1) \\( \frac { 25 } { 24 } \\) (3) \\( \frac { 7 } { 24 } \\)
(2) \\( \frac { 24 } { 7 } \\) (4) \\( \frac { 24 } { 25 } \\)

1. in the right triangle below, cos q=?

(1) \\( \frac { 12 } { 5 } \\) (3) \\( \frac { 12 } { 17 } \\)
(2) \\( \frac { 5 } { 12 } \\) (4) \\( \frac { 12 } { 13 } \\)

Explanation:

Step1: Recall trigonometric ratio definitions

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

Step2: Solve for \(\cos N\)

For \(\angle N\), the adjacent side is \(9\) and the hypotenuse is \(15\). So \(\cos N=\frac{9}{15}=\frac{3}{5}\)

Step3: Solve for \(\sin N\)

For \(\angle N\), the opposite side is \(12\) and the hypotenuse is \(15\). So \(\sin N=\frac{12}{15}=\frac{4}{5}\)

Step4: Solve for \(\tan N\)

For \(\angle N\), the opposite side is \(12\) and the adjacent side is \(9\). So \(\tan N=\frac{12}{9}=\frac{4}{3}\)

Step5: Solve for \(\sin P\)

For \(\angle P\), the opposite side is \(9\) and the hypotenuse is \(15\). So \(\sin P=\frac{9}{15}=\frac{3}{5}\)

Step6: Solve for \(\cos P\)

For \(\angle P\), the adjacent side is \(12\) and the hypotenuse is \(15\). So \(\cos P=\frac{12}{15}=\frac{4}{5}\)

Step7: Solve for \(\tan P\)

For \(\angle P\), the opposite side is \(9\) and the adjacent side is \(12\). So \(\tan P=\frac{9}{12}=\frac{3}{4}\)

Step8: Solve problem 7

For \(\tan A\) in \(\triangle ABC\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). For \(\angle A\), the opposite side is \(BC = 24\) and the adjacent side is \(AB=7\). So \(\tan A=\frac{24}{7}\) (Option 2)

Step9: Solve problem 8

First, find the hypotenuse of \(\triangle SRQ\) using the Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\), where \(a = 5\) and \(b = 12\). So \(SQ=\sqrt{5^{2}+12^{2}}=\sqrt{25 + 144}=\sqrt{169}=13\). Then \(\cos Q=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{12}{13}\) (Option 4)

Answer:

1. \(\frac{3}{5}\)
2. \(\frac{4}{5}\)
3. \(\frac{4}{3}\)
4. \(\frac{3}{5}\)
5. \(\frac{4}{5}\)
6. \(\frac{3}{4}\)
7. (2) \(\frac{24}{7}\)
8. (4) \(\frac{12}{13}\)