Question

Determine the length of the chord AB, to the nearest tenth of a centimeter.

Answer 1

Given:

`\begin{gathered} AD=BD=8.9m \\ DC=15.5m \\ \angle C=27.5\degree \end{gathered}`

Required:

To find the length of chord AB.

Step-by-step explanation:

Now consider the figure.

From the sine rule,

`\begin{gathered} (\sin C)/(BD)=(\sin B)/(DC) \\ \\ (\sin27.5)/(8.9)=(\sin B)/(15.5) \\ \\ \sin B=(\sin27.5)/(8.9)*15.5 \\ \\ B=126.46\degree \end{gathered}`

Now

`\begin{gathered} \angle A=180-126.46 \\ =53.54\degree \end{gathered}`
`\begin{gathered} \angle D=180-53.54-53.54 \\ =72.92\degree \end{gathered}`

Now from the sine rule

`\begin{gathered} (\sin D)/(AB)=(\sin B)/(AD) \\ \\ (\sin72.92)/(AB)=(\sin53.54)/(8.9) \\ \\ AB=(\sin72.29)/(\sin53.54)*8.9 \\ \\ AB=10.54m \end{gathered}`

Final Answer:

The length of chord AB is 10.54m.