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:

First, identify the sides relative to the 45° angle in right triangle \( \triangle ABC \) (right-angled at \( C \)):
- Hypotenuse: \( AB = 9 \)
- Opposite to \( 45^\circ \) (angle at \( A \)): \( BC \)
- Adjacent to \( 45^\circ \): \( AC \)

Step 1: Recall Trigonometric Ratios

- Sine: \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)
- Cosine: \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \)
- Tangent: \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \)

Step 2: Analyze Each Option

1. \( \sin(45^\circ) = \frac{BC}{9} \):
- \( \sin(45^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{BC}{9} \). This is correct.
2. \( \sin(45^\circ) = \frac{9}{BC} \):
- Sine is opposite over hypotenuse, not hypotenuse over opposite. Incorrect.
3. \( 9 \tan(45^\circ) = AC \):
- \( \tan(45^\circ) = \frac{BC}{AC} \), so \( AC = \frac{BC}{\tan(45^\circ)} \). Alternatively, \( \tan(45^\circ) = \frac{BC}{AC} \implies BC = AC \tan(45^\circ) \). Wait, let's re - check. Wait, angle at \( A \) is \( 45^\circ \), so opposite is \( BC \), adjacent is \( AC \). \( \tan(45^\circ)=\frac{BC}{AC}\implies BC = AC\tan(45^\circ) \). But also, from sine: \( BC = 9\sin(45^\circ) \), and since \( \tan(45^\circ) = 1 \), and \( \sin(45^\circ)=\cos(45^\circ)=\frac{\sqrt{2}}{2}\), and \( AC = 9\cos(45^\circ) \) (from cosine: \( \cos(45^\circ)=\frac{AC}{9}\implies AC = 9\cos(45^\circ) \)). Wait, \( \tan(45^\circ) = 1 \), and \( AC = 9\cos(45^\circ) \), \( BC = 9\sin(45^\circ) \), and since \( \sin(45^\circ)=\cos(45^\circ) \), \( AC = BC \). Now, \( 9\tan(45^\circ)=9\times1 = 9 \), but \( AC = 9\cos(45^\circ)\approx9\times0.707\approx6.36 \), so this is incorrect. Wait, no, wait: Wait, angle at \( A \) is \( 45^\circ \), so when we take angle at \( A \), the opposite side is \( BC \), adjacent is \( AC \). But also, in a right - isosceles triangle (since angle at \( A \) is \( 45^\circ \), angle at \( B \) is also \( 45^\circ \)), so \( AC = BC \). Let's use the cosine ratio: \( \cos(45^\circ)=\frac{AC}{9}\implies AC = 9\cos(45^\circ) \). And since \( \tan(45^\circ) = 1 \), and \( AC = BC \), and \( BC = 9\sin(45^\circ) \), then \( AC = 9\sin(45^\circ)=9\times\frac{\sqrt{2}}{2}\), and \( 9\tan(45^\circ)=9\times1 = 9 \), which is not equal to \( AC \) (since \( 9\times\frac{\sqrt{2}}{2}\approx6.36 eq9 \)). So this option is incorrect. Wait, maybe I made a mistake. Wait, no, let's re - express the tangent ratio. Wait, angle at \( A \): \( \tan(45^\circ)=\frac{BC}{AC}\implies AC=\frac{BC}{\tan(45^\circ)} \). Since \( \tan(45^\circ) = 1 \), \( AC = BC \). And from sine, \( BC = 9\sin(45^\circ) \), so \( AC = 9\sin(45^\circ) \). But \( 9\tan(45^\circ)=9\times1 = 9 \), and \( 9\sin(45^\circ)\approx6.36 eq9 \). So this option is incorrect. Wait, maybe the option is mis - written? Wait, no, let's check the fourth option.
4. \( (AC)\sin(45^\circ)=BC \):
- \( \sin(45^\circ)=\frac{BC}{AC}\implies BC = AC\sin(45^\circ) \). This is correct (since \( \sin(45^\circ)=\frac{BC}{AC}\implies BC = AC\times\sin(45^\circ) \)).
5. \( \cos(45^\circ)=\frac{BC}{9} \):
- Cosine is adjacent over hypotenuse, \( BC \) is opposite, not adjacent. Incorrect. Wait, no: Wait, adjacent to \( 45^\circ \) (angle at \( A \)) is \( AC \), opposite is \( BC \). So \( \cos(45^\circ)=\frac{AC}{9} \), not \( \frac{BC}{9} \). So this is incorrect. Wait, but earlier we thought the first option \( \sin(45^\circ)=\frac{BC}{9} \) is correct, and the fourth option \( (AC)\sin(45^\circ)=BC \) is also correct. Wait, let's re - verify the fourth option:
From \( \sin(45^\circ)=\frac{BC}{AC} \), cross - multiplying gives \( BC = AC\times\sin(45^\circ) \), so \( (AC)\sin(45^\circ)=BC \) is correct.
And the first option \( \sin(45^\circ)=\frac{BC}{9} \) is also correct because \( \sin(45^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{BC}{AB}=\frac{BC}{9} \).

Answer:

The correct equations are \( \sin(45^\circ) = \frac{BC}{9} \), \( (AC)\sin(45^\circ) = BC \)