Question

In a triangle ABC, a line BP is drawn from B so that P lies on the side AC. For the triangle, AP=27mm, BP=30mm, BC=82mm and the angle APB=76∘ apply. Determine the area of triangle ABC

Answer 1

SOLUTION

The diagram for this is shown below

From the diagram above, considering triangle BPC, let us find angle C

Using sine rule, we have

$\begin{gathered} \frac{sinC\degree}{30}=\frac{sin104\degree}{82} \\ sinC=\frac{30\times sin104\degree}{82} \\ C=sin^{-1}\frac{30sin104}{82} \\ C=20.79\degree \end{gathered}$

From the same triangle BPC to get angle B, we have

$\begin{gathered} B+P+C=180\degree \\ B+104+20.79=180 \\ B+124.79=180 \\ B=55.21\degree \end{gathered}$

From the same triangle BPC, using sine rule to get the side PC, which I called x, we have

$\begin{gathered} \frac{sinB}{x}=\frac{sinP}{82} \\ \frac{sin55.21\degree}{x}=\frac{sin104\degree}{82} \\ x=\frac{sin55.21\times82}{sin104\degree} \\ x=69.404mm \end{gathered}$

This makes the side AC to become

$27+69.404=96.404mm$

So, to get the area of the triangle ABC, we have that

$Area=\frac{1}{2}|AC|\times|BC|\times sinC$

Applying we have

$\begin{gathered} Area=\frac{1}{2}|96.404|\times|82|\times sin20.79\degree \\ =1402.94mm^2 \end{gathered}$

Hence the answer is approximately 1402.94 square-millimeters to the nearest hundredth