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} 9 tan(45°)=ac (ac)sin(45°)=bc cos(45°)=\frac{bc}{9}

Explanation:

Step1: Recall trigonometric ratios

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), and \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). For \(\triangle ABC\) with \(\angle A = 45^{\circ}\) and hypotenuse \(AB = 9\).
The side opposite to \(\angle A\) is \(BC\), and the side adjacent to \(\angle A\) is \(AC\).

Step2: Check \(\sin(45^{\circ})\)

By the formula \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\sin(45^{\circ})=\frac{BC}{AB}\). Since \(AB = 9\), \(\sin(45^{\circ})=\frac{BC}{9}\).

Step3: Check \(\cos(45^{\circ})\)

By the formula \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\cos(45^{\circ})=\frac{AC}{AB}\). Since \(AB = 9\), \(\cos(45^{\circ})=\frac{AC}{9}\), not \(\cos(45^{\circ})=\frac{BC}{9}\).

Step4: Check \(\tan(45^{\circ})\)

By the formula \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), \(\tan(45^{\circ})=\frac{BC}{AC}\). If we rewrite it, \(BC = AC\tan(45^{\circ})\). Also, from \(\sin(45^{\circ})=\frac{BC}{9}\), \(BC = 9\sin(45^{\circ})\) and from \(\cos(45^{\circ})=\frac{AC}{9}\), \(AC = 9\cos(45^{\circ})\). Since \(\sin(45^{\circ})=\cos(45^{\circ})\), \(BC = AC\).
The equation \(9\tan(45^{\circ})=AC\) is incorrect. From \(\tan(45^{\circ})=\frac{BC}{AC}=1\) (because \(\tan(45^{\circ}) = 1\)), \(BC = AC\). And from \(\cos(45^{\circ})=\frac{AC}{9}\), \(AC = 9\cos(45^{\circ})\).
The equation \((AC)\sin(45^{\circ})=BC\) is incorrect. Since \(BC = AC\) (because \(\tan(45^{\circ}) = 1=\frac{BC}{AC}\)), and \(\sin(45^{\circ})=\cos(45^{\circ})
eq1\) ( \(\sin(45^{\circ})=\cos(45^{\circ})=\frac{\sqrt{2}}{2}\)).
The equation \(\sin(45^{\circ})=\frac{9}{BC}\) is incorrect as per the ratio \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\).

Answer:

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