First problem: draw a line segment through the vertex R and a point C on PQ such that the segment is perpendicular to PQ, thus forming an altitude of the triangle. This new segment CR, the new segment CP, and PR form a 30-60-90 right triangle, where the measure of angle PCR is 90 degrees. CR then occurs with PR in a ratio of [tex]\sqrt3[/tex] to 2, so
[tex]\dfrac{CR}{PR}=\dfrac{CR}5=\dfrac{\sqrt3}2\implies CR=\dfrac{5\sqrt3}2[/tex]
and this can be considered the height of a triangle whose base is PQ. The area is then
[tex]\dfrac12\cdot6\cdot\dfrac{5\sqrt3}2=\dfrac{15\sqrt3}2\approx13[/tex]
Second problem: the sum of the interior angles of any triangle is 180, so the measure of the missing angle must be 180 - 20 - 35 = 125 degrees. By the law of sines, the length of the missing side (call it [tex]x[/tex]) satisfies
[tex]\dfrac{\sin20^\circ}9=\dfrac{\sin125^\circ}x\implies x\approx21.56[/tex]
Use Heron's formula to find the area of the triangle:
[tex]A=\sqrt{s(s-a)(s-b)(s-c)}[/tex]
where [tex]A[/tex] is the area, [tex]a,b,c[/tex] are the lengths of the sides, and [tex]s=\dfrac{a+b+c}2[/tex] is the half the perimeter of the triangle. Then the area is about [tex]A\approx54.62[/tex] square feet. Convert to square yards:
[tex]54.62\text{ ft}^2\cdot\left(\dfrac{1\text{ yd}}{3\text{ ft}}\right)^2\approx6.07\text{ yd}^2[/tex]
then round up to 7 square yards. At a price of $12.50 per square yard, the total price will be $87.50.
