Question

writing equations using trigonometric ratios
which equations could be used to solve for the unknown lengths of △abc? check all that apply.
□ \\(\sin(45^{\circ}) = \frac{bc}{9}\\)
□ \\(\sin(45^{\circ}) = \frac{9}{bc}\\)
□ \\(9 \tan(45^{\circ}) = ac\\)
□ \\((ac)\sin(45^{\circ}) = bc\\)
□ \\(\cos(45^{\circ}) = \frac{bc}{9}\\)

Explanation:

Step1: Identify sides for angle A

In $\triangle ABC$, $\angle C=90^\circ$, $\angle A=45^\circ$, hypotenuse $AB=9$.

• Opposite to $\angle A$: $BC$
• Adjacent to $\angle A$: $AC$

Step2: Apply sine ratio

Sine is opposite over hypotenuse.
$\sin(45^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{BC}{9}$

Step3: Apply tangent ratio

Tangent is opposite over adjacent.
$\tan(45^\circ)=\frac{BC}{AC} \implies AC \cdot \tan(45^\circ)=BC$
Rearrange from $\sin(45^\circ)=\frac{BC}{9}$: $9\sin(45^\circ)=BC$, substitute $BC=AC\tan(45^\circ)$:
$9\sin(45^\circ)=AC\tan(45^\circ)$, but $\tan(45^\circ)=1$, $\sin(45^\circ)=\frac{\sqrt{2}}{2}$, so $9 \cdot \frac{\sqrt{2}}{2}=AC$, which also matches rearranging $\sin(45^\circ)=\frac{BC}{9}$ and $BC=AC$ (45-45-90 triangle), so $9\tan(45^\circ)=AC$ is valid.

Step4: Rearrange sine for AC

From $\sin(45^\circ)=\frac{BC}{9}$, and $\cos(45^\circ)=\frac{AC}{9} \implies AC=9\cos(45^\circ)$. Substitute into $\sin(45^\circ)=\frac{BC}{AC}$:
$AC\sin(45^\circ)=BC$

Step5: Check cosine ratio

Cosine is adjacent over hypotenuse, so $\cos(45^\circ)=\frac{AC}{9}$, not $\frac{BC}{9}$.

Answer:

$\boldsymbol{\sin(45^{\circ}) = \frac{BC}{9}}$
$\boldsymbol{9 \tan(45^{\circ}) = AC}$
$\boldsymbol{(AC)\sin(45^{\circ}) = BC}$