Question

g for angle measures of right triangles which expressions can be used to find m∠bac? choose three correct answers.

Explanation:

Step1: Recall trigonometric - ratio definitions

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 $\angle BAC$ in right - triangle $ABC$ with right - angle at $C$, the opposite side to $\angle BAC$ is $BC = 9.8$ cm, the adjacent side is $AC=6.9$ cm, and the hypotenuse is $AB = 12$ cm.

Step2: Calculate $\tan\angle BAC$

$\tan\angle BAC=\frac{BC}{AC}=\frac{9.8}{6.9}$, so $\angle BAC=\tan^{- 1}(\frac{9.8}{6.9})$.

Step3: Calculate $\sin\angle BAC$

$\sin\angle BAC=\frac{BC}{AB}=\frac{9.8}{12}$, so $\angle BAC=\sin^{-1}(\frac{9.8}{12})$.

Step4: Calculate $\cos\angle BAC$

$\cos\angle BAC=\frac{AC}{AB}=\frac{6.9}{12}$, so $\angle BAC=\cos^{-1}(\frac{6.9}{12})$.

Answer:

$\tan^{-1}(\frac{9.8}{6.9})$, $\sin^{-1}(\frac{9.8}{12})$, $\cos^{-1}(\frac{6.9}{12})$