Question

math 2 unit 3 classwork soh cah toa directions - a.) find the value of x or θ in each problem below. round to the nearest tenth. b.) find the answer in the coloring sheet and color that piece according to \color\ that is associated to each set of problems. c.) you must show work to receive full credit for this worksheet! problem set #1 - color answers in this set blue problem set #2 - color answers in this set yellow problem set #3 - color answers in this set brown

Explanation:

Step1: Recall SOH - CAH - TOA

SOH: $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, CAH: $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, TOA: $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$.

Step2: For the first triangle in Problem Set #1 with sides 6 and 8

We want to find $\theta$. Using $\tan\theta=\frac{6}{8} = 0.75$. Then $\theta=\arctan(0.75)\approx36.9^{\circ}$.

Step3: For the second triangle in Problem Set #1 with opposite side 10 and adjacent side $x$ and $\theta = 39^{\circ}$

Using $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$, so $\tan39^{\circ}=\frac{10}{x}$. Then $x = \frac{10}{\tan39^{\circ}}\approx12.3$.

Step4: For the third triangle in Problem Set #1 with sides 9, 12 and 15

We want to find $\theta$. Using $\cos\theta=\frac{9}{15}=0.6$. Then $\theta=\arccos(0.6)\approx53.1^{\circ}$.

Step5: For the fourth triangle in Problem Set #1 with hypotenuse 7 and angle $54^{\circ}$ and opposite side $x$

Using $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, so $\sin54^{\circ}=\frac{x}{7}$. Then $x = 7\times\sin54^{\circ}\approx5.7$.

Step6: For the fifth triangle in Problem Set #1 with adjacent side 13.5 and angle $24^{\circ}$ and hypotenuse $x$

Using $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, so $\cos24^{\circ}=\frac{13.5}{x}$. Then $x=\frac{13.5}{\cos24^{\circ}}\approx14.8$.

Answer:

The values of $x$ and $\theta$ are found using SOH - CAH - TOA as shown above for each triangle. The full - set of calculations for all triangles in the worksheet should be done in a similar fashion following the trigonometric ratios.