Question

area of a triangle: law of sinesother tutors couldn’t solve so they referred me to different people!! please help i’m so desperate

Answer 1

For this problem, we use the sine law to compute the length of the missing sides:

`\frac{^{}VU}{\sin\text{ }\measuredangle\text{UWV}}=(WU)/(\sin \measuredangle WVU)`

Recalling that the interior angles of a triangle add up 180 degrees we get:

`\measuredangle WVU=180^(\circ)-26^(\circ)-35^(\circ)=119^(\circ)`

Substituting the angles and solving for the length of the missing sides we get:

`\begin{gathered} (WU)/(\sin 119^(\circ))=\frac{10\text{ yd}}{\sin 26^(\circ)}\Rightarrow VW=\sin 119^(\circ)(10yd)/(\sin 26^(\circ))\approx19.951\text{ yd} \\ \end{gathered}`

The height of the triangle with respect to the side WU is:

`h=10yd(\sin 35^(\circ))\approx5.735\text{ yd}`

Finally, using the formula for the area of a triangle:

`A=(bh)/(2)=\frac{19.951yd\cdot5.735\text{ yd}}{2}=57.2yd^2`

Answer: 57.2 square yards.