<p>Can anyone solve these 6 math triangle problem on this site.
I solve these but i don't if it is right. trigonometry is a hardest part that I struggle with.
really appreciate your help...
Science</a> - The everyday arithmetic</p>
<p>for number 17, we know that l represents the total side and the hypotenuse of the other triangle. Hence, l-lcos(theta) =h. then for 18 just plug in the values that they present in the question. For 5 the two right traingles are similar. Thus theta5 and theta 2 are 55. Then do 180= 70+2(theta1), and solve for theta one. Since you have theta 1 you can easily solve for the other two angles presented. Hope this helps!</p>
<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>