Question

QUESTIONEIn the diagram of AABC shown below, BC = 10 andAB 16.BсTo the nearest tenth of a degree, what is the measure of thelargest acute angle in the triangle?0.900O 38.70 32.004513

Answer 1

To find the angle θ, we are to apply the method of Cosine of angles.

`\begin{gathered} \cos \theta=(adjacent)/(hypotenuse) \\ \text{adjacent}=\text{ BC=10} \\ \text{hypotenuse}=AB=16 \\ \end{gathered}`
`\begin{gathered} \cos \theta=(10)/(16)=0.625 \\ \cos \theta=0.625 \\ \theta=\cos ^(-1)0.625 \\ \theta=51.3^0 \end{gathered}`

The total angle in a triangle=180.

So to get the remaining angle we need to add 90° to 51.3° and subtract them from 180°.

180°-(90°+51.3°)

180°-141.3°= 38.7°

An acute angle is an angle less than 90°, so therefore 51.3° is the largest acute angle.

Hence, option D is the right answer.