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°) = bc/9, sin(45°) = 9/bc, 9 tan(45°) = ac, (ac)sin(45°) = bc, cos(45°) = bc/9

Explanation:

Step1: Recall trigonometric - ratio definitions

In a right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, and $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. In right - triangle $ABC$ with $\angle A = 45^{\circ}$, the hypotenuse $AB = 9$.

Step2: Analyze $\sin(45^{\circ})$

Since $\sin(45^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}$ and the side opposite $\angle A$ is $BC$ and the hypotenuse is $AB = 9$, we have $\sin(45^{\circ})=\frac{BC}{9}$, so the first equation is correct.

Step3: Analyze $\sin(45^{\circ})=\frac{9}{BC}$

This is incorrect because by the definition of sine, $\sin(45^{\circ})=\frac{BC}{9}$, not $\frac{9}{BC}$.

Step4: Analyze $9\tan(45^{\circ})=AC$

We know that $\tan(45^{\circ}) = 1$. Also, $\tan(45^{\circ})=\frac{BC}{AC}$. From $\sin(45^{\circ})=\frac{BC}{9}$, we get $BC = 9\sin(45^{\circ})$. And $\tan(45^{\circ})=\frac{BC}{AC}=1$ implies $BC = AC$. But $9\tan(45^{\circ})=9
eq AC$ (since $AC = BC=9\sin(45^{\circ})$), so this equation is incorrect.

Step5: Analyze $(AC)\sin(45^{\circ})=BC$

Since $\tan(45^{\circ}) = 1$, $AC = BC$. And $\sin(45^{\circ})=\frac{BC}{9}$, so $(AC)\sin(45^{\circ})=BC$ is incorrect.

Step6: Analyze $\cos(45^{\circ})=\frac{BC}{9}$

We know that $\cos(45^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}$. The side adjacent to $\angle A$ is $AC$, not $BC$. So $\cos(45^{\circ})=\frac{AC}{9}$, not $\frac{BC}{9}$, and this equation is incorrect.

Answer:

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