Question

In ∆ABC, a= 420° cm, b=620cm and c=410 cm. find the area of ∆ABC to the nearest square centimeter.

Answer 1

Let's make a diagram

Let's find angle C using the law of cosines

$\begin{gathered} c^2=a^2+b^2-2ab\cdot\cos C \\ 410^2=420^2+620^2-2\cdot420\cdot620\cdot\cos C \\ 168100=176400+384400-520800\cdot\cos C \\ 168100=560800-520800\cdot\cos C \\ C=\cos ^{-1}(\frac{168100-560800}{-520800}_{}) \\ C=41.06 \end{gathered}$

Then, we use the following formula to find the area

$\begin{gathered} A=\frac{1}{2}ab\sin C \\ A=\frac{1}{2}\cdot420\cdot620\cdot\sin 41.06 \\ A=85,522(cm)^2 \end{gathered}$

However, if we use all the decimal numbers of the angle, we get an area of 85,520 square centimeters.