Question

for $\triangle abc$, which two relationships are true?$square$ $cos(\theta) = \frac{ac}{bc} = sin(\theta)LXB0square$ $sin(\theta) = \frac{bc}{ac} = cos(90 - \theta)LXB1square$ $cos(\theta) = \frac{ab}{ac} = sin(90 - \theta)$$square$ $sin(\theta) = \frac{ab}{ac} = cos(90 - \theta)$

Explanation:

Step1: Define sides for angle $\theta$

In right $\triangle ABC$, $\angle B=90^\circ$, $\angle A=\theta$:

• Adjacent to $\theta$: $AB$
• Opposite to $\theta$: $BC$
• Hypotenuse: $AC$

Step2: Recall trigonometric identities

$\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$, and $\cos(90^\circ-\theta)=\sin(\theta)$, $\sin(90^\circ-\theta)=\cos(\theta)$

Step3: Evaluate each option

1. $\cos(\theta)=\frac{AC}{BC}=\sin(\theta)$: $\cos(\theta)=\frac{AB}{AC}$, false.
2. $\sin(\theta)=\frac{AB}{AC}=\cos(\theta)$: $\sin(\theta)=\frac{BC}{AC}$, false.
3. $\sin(\theta)=\frac{BC}{AC}=\cos(90-\theta)$: $\sin(\theta)=\frac{BC}{AC}$, $\cos(90-\theta)=\sin(\theta)$, true.
4. $\cos(\theta)=\frac{AB}{BC}=\sin(90-\theta)$: $\cos(\theta)=\frac{AB}{AC}$, false.
5. $\cos(\theta)=\frac{AB}{AC}=\sin(90-\theta)$: $\cos(\theta)=\frac{AB}{AC}$, $\sin(90-\theta)=\cos(\theta)$, true.
6. $\sin(\theta)=\frac{AB}{AC}=\cos(90-\theta)$: $\sin(\theta)=\frac{BC}{AC}$, false.

Answer:

$\sin(\theta) = \frac{BC}{AC} = \cos(90 - \theta)$
$\cos(\theta) = \frac{AB}{AC} = \sin(90 - \theta)$