Question

Find the area of this triangle. Round to the nearest tenth. PLEASE HELP

Answer 1

74.8 m²

Explanation:

`\boxed{\begin{minipage}{7.6 cm}\underline{Area of a triangle} \\\\$A=(1)/(2)ab \sin C$\\\\where:\\ \phantom{ww}$\bullet$ $C$ is the angle. \\ \phantom{ww}$\bullet$ $a$ and $b$ are the sides enclosing the angle. \\\end{minipage}}`

From inspection of the given triangle:

• C = 115°
• a = 11 m
• b = 15 m

Substitute the values into the formula and solve for area:

`\implies A=(1)/(2) \cdot 11 \cdot 15 \cdot \sin 115^(\circ)`

`\implies A=(165)/(2) \sin 115^(\circ)`

`\implies A=74.77039243`

`\implies A=74.8\; \sf m^2\;(nearest\;tenth)`

Therefore, the area of the triangle is 74.8 m² rounded to the nearest tenth.