Question

which equations could be used to solve for the unknown lengths of △abc? check all that apply

Explanation:

Step1: Recall trigonometric ratios

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\).
For \(\angle A = 45^{\circ}\), hypotenuse \(AB = 8\).
The side opposite to \(\angle A\) is \(BC\), and the side adjacent to \(\angle A\) is \(AC\).

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

By the definition of sine, \(\sin(45^{\circ})=\frac{BC}{AB}\). Since \(AB = 8\), we have \(\sin(45^{\circ})=\frac{BC}{8}\), which can be rewritten as \(BC = 8\sin(45^{\circ})\). Also, from \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), if we consider \(\sin(45^{\circ})=\frac{BC}{8}\), cross - multiplying gives \(8\sin(45^{\circ})=BC\).
Since \(\angle A=45^{\circ}\), \(\triangle ABC\) is a \(45 - 45-90\) triangle, so \(AC = BC\). And \(\sin(45^{\circ})=\frac{AC}{AB}\) (because in a right - triangle, for an angle \(\theta\), \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), and if we swap the role of opposite and adjacent (since \(AC = BC\) in a \(45 - 45-90\) triangle), \((AC)\sin(45^{\circ})=BC\) (because \(AC = BC\) and \(\sin(45^{\circ}) = 1\) in the context of equality of the two legs of a \(45 - 45-90\) triangle, more precisely, using \(\sin(45^{\circ})=\frac{BC}{AB}\) and \(AB=\sqrt{AC^{2}+BC^{2}}\), when \(AC = BC\), \(AB=\sqrt{2}BC\), and \(\sin(45^{\circ})=\frac{1}{\sqrt{2}}\), \(BC = AB\sin(45^{\circ})\) and \(AC = BC\)).

Step3: Analyze other options

• For \(\sin(45^{\circ})=\frac{8}{BC}\), by the definition \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), the hypotenuse is the longest side. Here, if \(\sin(45^{\circ})=\frac{8}{BC}\), then \(BC\) would be the hypotenuse, which is wrong as \(AB = 8\) is the hypotenuse.
• For \(8\tan(45^{\circ})=AC\), \(\tan(45^{\circ})=\frac{BC}{AC}\), and since \(AC = BC\) in a \(45 - 45-90\) triangle, \(\tan(45^{\circ}) = 1\), but \(AB\tan(45^{\circ})

eq AC\) (because \(AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{2}AC\)).

• For \(\cos(45^{\circ})=\frac{BC}{8}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), the adjacent side to \(\angle A\) is \(AC\), not \(BC\).

Answer:

\(\sin(45^{\circ})=\frac{BC}{8}\), \((AC)\sin(45^{\circ})=BC\)