<p>For each one these, there should be enough equations to clearly describe the systems.</p>
<p>In 5, these are
t1 + t4 = 90
t4 + t5 = 90
2*t1 = 110
t2 + t5 = 110
t2 + t3 = 90
t1 + t3 = 90</p>
<p>(You really only need 5 of the 6)</p>
<p>In 6, these are
A = D
C = 90
B = 90 - D
B + 37 = 90</p>
<p>There are exactly 4 equations here, so this system should be exactly solvable</p>
<p>In 17, if you remember the moniker about cos(theta) = adjacent/hypothenuse of the corresponding triangle, then we have a right triangle with hypothenuse l and adjacent side l-h. This then gives us an expression for theta as</p>
<p>cos(theta) = l/(l-h) // note that only the denominator on the RHS contains the h
l - h = l/cos(theta)
h = l - l/cos(theta)</p>
<p>In 18, this is just 6*(1 - secant(40-degrees))</p>
<p>For 1, observe that if you have two squares with area A, and you tape them together, the total area of whatever shape you make from that (as long as it is not folded, which implies that there’s an extra dimension) is 2A.</p>
<p>There are two shapes, you need to find the area of both and add them.</p>
<p>One shape is a rectangle, the other is a trapezoid, both have height of 6m.</p>
<p>For 2, if we assume that the top 5m of the shape is half of a circle, then our two shapes are half of a circle and a rectangle.</p>