Question

In trIangle OPQ, p = 3.4 cm, q = 3.2 cm and angle 0=99º. Find the area of Triangle OPQ, to the nearest10th of a square centimeter.

Answer 1

we know that

The area of a triangle applying the law of sines is equal to

`A=(1)/(2)\cdot p\cdot q\cdot\sin (O)`

substitute the given values

`\begin{gathered} A=(1)/(2)\cdot3.4\cdot3.2\cdot\sin (99^o) \\ A=5.37\text{ cm\textasciicircum{}2} \end{gathered}`

the answer is

5.37 square centimeters