Question

6.

1. 16 in, \\(\sqrt{37}\\) in, x
2. 15, 26°, x
3. v, 1, 60°, u
4. m, n, 45°, \\(5\sqrt{2}\\)

Explanation:

Problem 6

Step1: Identify trigonometric ratio

We know the hypotenuse ($17$) and need side $x$, adjacent to $59^\circ$. Use cosine: $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$

Step2: Solve for $x$

$x = 17 \times \cos(59^\circ)$
$x \approx 17 \times 0.5150 = 8.755$

Problem 7

Step1: Apply Pythagorean Theorem

For right triangle: $a^2 + b^2 = c^2$, where $c=16$, $b=\sqrt{37}$, solve for $x$.

Step2: Rearrange and calculate

$x^2 = 16^2 - (\sqrt{37})^2 = 256 - 37 = 219$
$x = \sqrt{219} \approx 14.799$

Problem 8

Step1: Identify trigonometric ratio

We know hypotenuse ($15$) and need side $x$, opposite to $26^\circ$. Use sine: $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$

Step2: Solve for $x$

$x = 15 \times \sin(26^\circ)$
$x \approx 15 \times 0.4384 = 6.576$

Problem 9

Step1: Use trigonometry for $v$

$v$ is opposite $60^\circ$, adjacent side is $1$. $\tan(60^\circ)=\frac{v}{1}$
$v = \tan(60^\circ) = \sqrt{3} \approx 1.732$

Step2: Use trigonometry for $u$

$u$ is hypotenuse, adjacent side is $1$. $\cos(60^\circ)=\frac{1}{u}$
$u = \frac{1}{\cos(60^\circ)} = \frac{1}{0.5} = 2$

Problem 10

Step1: Identify 45-45-90 triangle properties

Legs are equal, hypotenuse $= \text{leg} \times \sqrt{2}$. Given hypotenuse $5\sqrt{2}$, so leg $n=5$.

Step2: Calculate hypotenuse $m$

$m = 5\sqrt{2}$ (matches given hypotenuse, or use Pythagorean theorem: $m^2 = 5^2 + 5^2 = 50$, $m=\sqrt{50}=5\sqrt{2}$)

Answer:

1. $x \approx 8.76$
2. $x = \sqrt{219} \approx 14.80$
3. $x \approx 6.58$
4. $v = \sqrt{3} \approx 1.73$, $u=2$
5. $n=5$, $m=5\sqrt{2}$

Method Selections:

1. SOH CAH TOA
2. Pythagorean Theorem
3. SOH CAH TOA
4. SOH CAH TOA
5. 45-45-90