Which expressions can be used to find m∠bac? select three options. cos−1(startfraction 6.9 over 12 endfraction) cos−1(startfraction 9.8 over 12 endfraction) sin−1(startfraction 6.9 over 12 endfraction) sin−1(startfraction 9.8 over 12 endfraction) tan−1(startfraction 6.9 over 9.8 endfraction)

The correct answers are A, D and E

Step-by-step explanation:

Video explanation

The previous answer said F but it's actually E :)

You're welcome

A, D, F

Step-by-step explanation: Angle BAC is angle A in triangle ABC. Use the definitions of the trigonometric functions:

\cos \angle A=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{AC}{AB}=\dfrac{6.9}{12}\\ \\\sin \angle A=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{BC}{AB}=\dfrac{9.8}{12}\\ \\\tan \angle A=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{BC}{AC}=\dfrac{9.8}{6.9}\\ \\\cot \angle A=\dfrac{\text{adjacent leg}}{\text{opposite leg}}=\dfrac{AC}{BC}=\dfrac{6.9}{9.8}

Now

\angle A=\cos^{-1}\left(\dfrac{6.9}{12}\right)\\ \\\angle A=\sin^{-1}\left(\dfrac{9.8}{12}\right)\\ \\\angle A=\tan^{-1}\left(\dfrac{9.8}{6.9}\right)\\ \\\angle A=\cot^{-1}\left(\dfrac{6.9}{9.8}\right)

So, correct options are A, D and F