Question

1. which segment is the hypotenuse?

a. $overline{qj}$
b. $overline{gj}$
c. $overline{gq}$

1. which segment is opposite to $angle b$

a. $overline{aw}$
b. $overline{wb}$
c. $overline{ab}$

1. solve for $x$. round to the nearest tenth, if necessary.

$cos18=\frac{x}{52}$
$x = 52cos18$
$x = 34.3$

1. solve for $x$. round to the nearest tenth, if necessary.

$sin68=\frac{77}{x}$
$x=\frac{77}{sin68}$

1. solve for $x$. round to the nearest tenth of a degree, if necessary.

$sin^{-1}=\frac{opp}{hyp}$
$sin^{-1}=\frac{9}{18}$
$x = 0.52$

Explanation:

Step1: Identify hypotenuse in right - triangle

In a right - triangle, the hypotenuse is the side opposite the right - angle. In right - triangle $QJG$ with right - angle at $Q$, the side opposite $\angle Q$ is $\overline{GJ}$.

Step2: Identify opposite side to an angle

In right - triangle $ABW$ with right - angle at $W$, the side opposite $\angle B$ is $\overline{AW}$.

Step3: Solve for $x$ in $\cos18^{\circ}=\frac{x}{52}$

We know that $x = 52\cos18^{\circ}$. Using a calculator, $\cos18^{\circ}\approx0.9511$, so $x = 52\times0.9511\approx49.4$.

Step4: Solve for $x$ in $\sin68^{\circ}=\frac{77}{x}$

We can rewrite the equation as $x=\frac{77}{\sin68^{\circ}}$. Since $\sin68^{\circ}\approx0.9272$, then $x=\frac{77}{0.9272}\approx83.0$.

Step5: Solve for $x$ in $\sin^{-1}(\frac{9}{18}) = x$

First, simplify $\frac{9}{18}=\frac{1}{2}$. Then $\sin^{-1}(\frac{1}{2})=30^{\circ}$.

Answer:

1. B. $\overline{GJ}$
2. A. $\overline{AW}$
3. $x\approx49.4$
4. $x\approx83.0$
5. $x = 30^{\circ}$