Question

which equations could be used to solve for the unknown lengths of △abc? check all that apply.
□ sin(45°) = \frac{bc}{9}
□ sin(45°) = \frac{9}{bc}
□ sin(45°) = \frac{ac}{9}
□ 9 tan(45°) = ac
□ (ac)sin(45°) = bc
□ cos(45°) = \frac{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 right - angle at $C$, $\angle A = 45^{\circ}$ and hypotenuse $AB = 9$. Let $BC$ be the side opposite to $\angle A$ and $AC$ be the side adjacent to $\angle A$.

Step2: Analyze each option

• For $\cos(45^{\circ})=\frac{BC}{9}$: Since $\cos(45^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}$, and the adjacent side to $\angle A$ is $AC$, not $BC$, this is incorrect.
• For $(AC)\sin(45^{\circ})=BC$: We know that $\sin(45^{\circ})=\frac{BC}{AB}$, and $AB = 9$. Also, in a $45 - 45-90$ triangle, $AC = BC$. But from $\sin(45^{\circ})=\frac{BC}{9}$, we can rewrite it as $BC = 9\sin(45^{\circ})$. And since $\sin(45^{\circ})=\frac{BC}{AB}$, cross - multiplying gives $AB\sin(45^{\circ})=BC$. Since $AB = 9$, we have $9\sin(45^{\circ})=BC$. Also, since $AC = BC$ in a $45 - 45-90$ triangle, $AC\sin(45^{\circ})=BC$ is correct.
• For $9\tan(45^{\circ})=AC$: Since $\tan(45^{\circ}) = 1$, then $9\tan(45^{\circ})=9$. And in a $45 - 45-90$ triangle with hypotenuse $AB = 9$, $AC=\frac{9}{\sqrt{2}}$, so this is incorrect.
• For $\sin(45^{\circ})=\frac{BC}{9}$: Since $\sin(45^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}$ and the opposite side to $\angle A$ is $BC$ and hypotenuse $AB = 9$, this is correct.
• For $\sin(45^{\circ})=\frac{9}{BC}$: This is incorrect as $\sin(45^{\circ})=\frac{BC}{9}$ (opposite over hypotenuse).

Answer:

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