Question

directions: give each trig ratio as a fraction in simplest form.
1.
directions: solve for x. round to the nearest tenth.
2.
3.
4.
5.
6.
7.

Explanation:

Step1: Find side PR in first triangle

Use Pythagorean theorem $a^{2}+b^{2}=c^{2}$. Let $PR = a$, $PQ=14$, $QR = 50$. Then $a=\sqrt{50^{2}-14^{2}}=\sqrt{(50 + 14)(50 - 14)}=\sqrt{64\times36}=48$.

Step2: Calculate trig - ratios for first triangle

Recall $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$.

• $\sin Q=\frac{PR}{QR}=\frac{48}{50}=\frac{24}{25}$
• $\cos Q=\frac{PQ}{QR}=\frac{14}{50}=\frac{7}{25}$
• $\tan Q=\frac{PR}{PQ}=\frac{48}{14}=\frac{24}{7}$
• $\sin R=\frac{PQ}{QR}=\frac{14}{50}=\frac{7}{25}$
• $\cos R=\frac{PR}{QR}=\frac{48}{50}=\frac{24}{25}$
• $\tan R=\frac{PQ}{PR}=\frac{14}{48}=\frac{7}{24}$

Step3: Solve for $x$ in question 2

We know $\tan48^{\circ}=\frac{x}{17}$, so $x = 17\times\tan48^{\circ}\approx17\times1.1106\approx18.9$.

Step4: Solve for $x$ in question 3

We know $\cos67^{\circ}=\frac{x}{29}$, so $x = 29\times\cos67^{\circ}\approx29\times0.3907\approx11.3$.

Step5: Solve for $x$ in question 4

We know $\tan29^{\circ}=\frac{12}{x}$, so $x=\frac{12}{\tan29^{\circ}}\approx\frac{12}{0.5543}\approx21.6$.

Step6: Solve for $x$ in question 5

We know $\sin16^{\circ}=\frac{x}{37}$, so $x = 37\times\sin16^{\circ}\approx37\times0.2756\approx10.2$.

Step7: Solve for $x$ in question 6

We know $\tan58^{\circ}=\frac{x}{22}$, so $x = 22\times\tan58^{\circ}\approx22\times1.6003\approx35.2$.

Step8: Solve for $x$ in question 7

We know $\tan51^{\circ}=\frac{x}{15}$, so $x = 15\times\tan51^{\circ}\approx15\times1.2349\approx18.5$.

Answer:

1. $\sin Q=\frac{24}{25}$, $\cos Q=\frac{7}{25}$, $\tan Q=\frac{24}{7}$, $\sin R=\frac{7}{25}$, $\cos R=\frac{24}{25}$, $\tan R=\frac{7}{24}$
2. $x\approx18.9$
3. $x\approx11.3$
4. $x\approx21.6$
5. $x\approx10.2$
6. $x\approx35.2$
7. $x\approx18.5$