For this problem, we use the sine law to compute the length of the missing sides:
[tex]\frac{^{}VU}{\sin\text{ }\measuredangle\text{UWV}}=\frac{WU}{\sin \measuredangle WVU}[/tex]
Recalling that the interior angles of a triangle add up 180 degrees we get:
[tex]\measuredangle WVU=180^{\circ}-26^{\circ}-35^{\circ}=119^{\circ}[/tex]
Substituting the angles and solving for the length of the missing sides we get:
[tex]\begin{gathered} \frac{WU}{\sin 119^{\circ}}=\frac{10\text{ yd}}{\sin 26^{\circ}}\Rightarrow VW=\sin 119^{\circ}\frac{10yd}{\sin 26^{\circ}}\approx19.951\text{ yd} \\ \end{gathered}[/tex]
The height of the triangle with respect to the side WU is:
[tex]h=10yd(\sin 35^{\circ})\approx5.735\text{ yd}[/tex]
Finally, using the formula for the area of a triangle:
[tex]A=\frac{bh}{2}=\frac{19.951yd\cdot5.735\text{ yd}}{2}=57.2yd^2[/tex]
Answer: 57.2 square yards.
