Question

select all the correct answers. for $\triangle abc$, which two relationships are true? options with trigonometric ratios and a right triangle diagram with angle $\theta$ at $a$ and angle $(90 - \theta)$ at $c$, right - angled at $b$

Explanation:

Step1: Recall Trigonometric Ratios

In right triangle \( \triangle ABC \) with right angle at \( B \), for angle \( \theta \) at \( A \):

• \( \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{BC}{AC} \)
• \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{AB}{AC} \)

Also, recall the co - function identity: \( \sin(90^{\circ}-\alpha)=\cos(\alpha) \) and \( \cos(90^{\circ}-\alpha)=\sin(\alpha) \). For \( \triangle ABC \), angle at \( C \) is \( 90^{\circ}-\theta \).

Step2: Analyze Each Option

• Option 1: \( \cos(\theta)=\frac{AB}{AC} \), and \( \sin(90 - \theta)=\cos(\theta) \) (by co - function identity, since \( 90-\theta \) is angle at \( C \), \( \sin(90 - \theta)=\frac{AB}{AC}=\cos(\theta) \)). So this is correct.
• Option 2: \( \sin(\theta)=\frac{AB}{AC} \) is wrong because \( \sin(\theta)=\frac{BC}{AC} \).
• Option 3: \( \cos(\theta)=\frac{AC}{BC} \) is wrong, \( \cos(\theta)=\frac{AB}{AC} \).
• Option 4: \( \sin(\theta)=\frac{AB}{AC} \) is wrong, \( \sin(\theta)=\frac{BC}{AC} \).
• Option 5: \( \sin(\theta)=\frac{BC}{AC} \) and \( \cos(90 - \theta)=\sin(\theta) \) (by co - function identity, \( \cos(90 - \theta)=\sin(\theta) \) since \( 90 - \theta \) is angle at \( C \), \( \cos(90 - \theta)=\frac{BC}{AC}=\sin(\theta) \)). So this is correct.
• Option 6: \( \cos(\theta)=\frac{AB}{BC} \) is wrong, \( \cos(\theta)=\frac{AB}{AC} \). And \( \sin(90 - \theta)=\cos(\theta) \), but \( \cos(\theta)

eq\frac{AB}{BC} \).

Answer:

The correct options are:

• \( \cos(\theta)=\frac{AB}{AC}=\sin(90 - \theta) \) (first option)
• \( \sin(\theta)=\frac{BC}{AC}=\cos(90 - \theta) \) (fifth option)