Question

select all of the equations that can be used to determine the value of w
$\tan\left(x\
ight)=\frac{9}{w}$
$9^2 + w^2 = (\sqrt{65})^2$
$w = \sqrt{9^2 + (65)^2}$
$\sin\left(90 - x\
ight)=\frac{9}{w}$
$\tan\left(90 - x\
ight)=\frac{9}{w}$

Explanation:

Step1: Analyze right triangle sides

In right $\triangle CRY$, $\angle R=90^\circ$:

• Adjacent to $\angle x$: $w$
• Opposite to $\angle x$: $9$
• Hypotenuse: $\sqrt{65}$
• $\angle (90-x)$ is $\angle Y$, opposite side $9$, adjacent side $w$

Step2: Check tangent of $\angle x$

$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$, so $\tan(x)=\frac{9}{w}$. This is valid.

Step3: Check Pythagorean theorem

Pythagorean theorem: $a^2+b^2=c^2$. Substitute values:
$$9^2 + w^2 = (\sqrt{65})^2$$
This is valid.

Step4: Check incorrect side formula

$w=\sqrt{9^2+(65)^2}$ uses hypotenuse as a leg, violates Pythagoras. Invalid.

Step5: Check sine of $\angle (90-x)$

$\sin(90-x)=\sin(Y)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{9}{\sqrt{65}}$, not $\frac{9}{w}$. Invalid.

Step6: Check tangent of $\angle (90-x)$

$\tan(90-x)=\tan(Y)=\frac{\text{opposite}}{\text{adjacent}}=\frac{9}{w}$. This is valid.

Answer:

• $\boldsymbol{\tan(x)=\frac{9}{w}}$
• $\boldsymbol{9^2 + w^2 = (\sqrt{65})^2}$
• $\boldsymbol{\tan(90-x)=\frac{9}{w}}$