Question

select all the equations that can be used to determine the value of w.$sin(90 - x) = \frac{9}{w}$$w = sqrt{9^2 + (125)^2}$$\tan(90 - x) = \frac{9}{w}$$\tan(x) = \frac{9}{w}$

Explanation:

Step1: Analyze right triangle properties

In right $\triangle CRH$, $\angle R=90^\circ$, $\angle C=x$, so $\angle H=90^\circ - x$. Sides: $CR=9$, $RH=w$, $CH=\sqrt{125}$.

Step2: Check each trigonometric equation

For $\sin(90-x)=\frac{9}{w}$:

$\sin(90-x)=\cos(x)=\frac{\text{adjacent to }x}{\text{hypotenuse}}=\frac{9}{\sqrt{125}}$. This does not equal $\frac{9}{w}$, so invalid.

For $w=\sqrt{9^2+(125)^2}$:

Pythagorean theorem: $\text{hypotenuse}^2=\text{leg}_1^2+\text{leg}_2^2$, so $(\sqrt{125})^2=9^2+w^2$, rearranged to $w=\sqrt{125 - 9^2}$. The given formula uses $125$ instead of $(\sqrt{125})^2$, so invalid.

For $\tan(90-x)=\frac{9}{w}$:

$\tan(90-x)=\cot(x)=\frac{\text{adjacent to }x}{\text{opposite to }x}=\frac{9}{w}$. This matches, so valid.

For $\tan(x)=\frac{9}{w}$:

$\tan(x)=\frac{\text{opposite to }x}{\text{adjacent to }x}=\frac{w}{9}$. This does not equal $\frac{9}{w}$, so invalid.

Answer:

$\boldsymbol{\tan(90 - x) = \frac{9}{w}}$