Here are two questions that look similar, but require different methods for solving.
The first, a problem involving areas, perimeters and some basic trigonometry.
The second involving equations and Trial and Improvement.
Solutions
Problem 1
The perimeter is \(30\), which means the area is also \(30\). The perpendicular height is the area divided by base width, \(30 \div 10 = 3\). The acute angle is \(x\) where \(sin(x) = \frac{3}{4}\) so \(x=\sin^{-1}(\frac{3}{4})=36.87^{\circ}\) resulting in the obtuse angle \(180-36.87=143.13^{\circ}\)
Problem 2
The perpendicular height of the parallelogram is \(xsin(x)\) so the area is \(10xsin(x)\). The perimeter is \(2x+20\) so we can form the equation \[2x+20=10xsin(x)\]Divide through by 2 and rearrange to get \[5xsin(x)-x=10\] One of the \(x\)'s is trapped inside the sine function and the other two are not. We don't have an analytical solution for this and have to resort to numerical methods. Through some trial and improvement, you can find the angle to be \(18.09^{\circ}\)
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