Question

a painter needs to cover a triangular region 60 meters by 68 meters by 71 meters. a can of can cover 70 square meters.. how many cans will be needed??

Answer 1

You want to set it up as a triangle and solve for the angles so you can determine the base and height of the triangle:


The law of cosines says \[c ^{2} = b^{2} + a^{2} - 2bacos(C)\]

Rearranged, you can solve for C:

\[C = \cos^{-1}\left[\frac{60^{2} + 68^{2} - 71^{2}}{2(60)(68)} \right] = 67.04^{o}\]

Then using the law of sines, you can solve for one of the other angles:

\[\frac{sinA}{60} = \frac{\sin67^{o}}{71}\]

\[A = 51.1^{o}\]

Then B is 61.86 because 180 - 67.04 - 51.1 = 61.86

Then solve for the base of the triangle:
\[Base = sinA \times b = \sin(51.1^{o}) \times 68 = 52.92m\]

The area of the triangle is 1/2 base times height, so:
\[Area = \frac{52.89 \times 71}{2} = 1877.595m^{2}\]

Divided by 70, that's 26.823 paint buckets.