Question

QuestionIn ABC, we are told that b = 5, B = 19°, and C = 66º. Solve for a and c. Round to the nearest tenth.

Answer 1

To solve this question we will use the sine Law:

`(c)/(\sin C)=(b)/(\sin B)=(a)/(\sin A)\text{.}`

To find the measure of angle A, recall that the interior angles of a triangle add up to 180 degrees, therefore:

`\measuredangle A+\measuredangle B+\measuredangle C=180^(\circ).`

From the given data we know that:

`\begin{gathered} \measuredangle B=19^(\circ), \\ \measuredangle C=66^(\circ). \end{gathered}`

Therefore:

`\measuredangle A+19^(\circ)+66^(\circ)=180^(\circ).`

Solving the above equation for angle A we get:

`\begin{gathered} \measuredangle A=180^(\circ)-19^(\circ)-66^(\circ), \\ \measuredangle A=96^(\circ). \end{gathered}`

Now, from the given data we get that b=5, therefore:

`\begin{gathered} (a)/(\sin96^(\circ))=(5)/(\sin 19^(\circ)), \\ (c)/(\sin 66^(\circ))=(5)/(\sin19^(\circ))\text{.} \end{gathered}`

Solving the above equations for a and c respectively we get:

`\begin{gathered} (a)/(\sin96^(\circ))*\sin 96^(\circ)=(5)/(\sin19^(\circ))*\sin 96^(\circ), \\ a=(5\sin96^(\circ))/(\sin19^(\circ)), \\ (c)/(\sin66^(\circ))*\sin 66^(\circ)=(5)/(\sin19^(\circ))*\sin 66^(\circ) \\ c=(5\sin66^(\circ))/(\sin19^(\circ)), \end{gathered}`

Therefore:

`\begin{gathered} a\approx15.3, \\ c\approx14.0. \end{gathered}`

Answer:

`\begin{gathered} a\approx15.3, \\ c\approx14.0. \end{gathered}`