which expressions can be used to find m∠bac? select three options.\n□cos⁻¹(6.9/12)\n□cos⁻¹(9.8/12)\n□sin⁻¹(6…

which expressions can be used to find m∠bac? select three options.\n□cos⁻¹(6.9/12)\n□cos⁻¹(9.8/12)\n□sin⁻¹(6.9/12)\n□sin⁻¹(9.8/12)\n□tan⁻¹(9.8/6.9)

which expressions can be used to find m∠bac? select three options.\n□cos⁻¹(6.9/12)\n□cos⁻¹(9.8/12)\n□sin⁻¹(6.9/12)\n□sin⁻¹(9.8/12)\n□tan⁻¹(9.8/6.9)

Answer

Explanation:

Step1: Recall trigonometric - ratio definitions

In right - triangle (ABC) with right - angle at (C), (\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}), (\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}), and (\tan\theta=\frac{\text{opposite}}{\text{adjacent}}) for an acute angle (\theta). For (\angle BAC), the adjacent side to (\angle BAC) is (AC = 6.9) cm, the opposite side is (BC=9.8) cm, and the hypotenuse is (AB = 12) cm.

Step2: Calculate cosine of (\angle BAC)

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

Step3: Calculate sine of (\angle BAC)

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

Step4: Calculate tangent of (\angle BAC)

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

Answer:

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