A triangular plot of land has sides of lengths 420 feet 350 feet and 180 feet. Approximate the smallest angle between side of the triangle. I need a sketch with it and final answer rounded to the tenths place.
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Let the given sides be denoted as a = 420 ft, b = 350 ft and c= 180 ft.
And the angles opposite to these sides be A, B and C, (degrees) respectively.
These angles can be determined using the cosine rule and the fact that sum of angles in a triangle is always 180 degrees.
i.e., A + B + C =180
and a^2 = b^2 + c^2 - 2bc cosA
i.e., cosA = (b^2 + c^2-a^2)/2bc = (350^2 + 180^2 - 420^2)/(2x350x180) = -0.1706
or, A = cos^(-1) (-0.1706) = 80.1753 degrees
similarly, b^2 = a^2 + c^2 - 2ac cos B
or, cosB = (a^2+c^2-b^2)/(2ac) = (420^2 + 180^2-350^2)/(2x420x180) = 0.57076
or, B = 55.1963 degrees
and C = 180-A-B = 44.6284 degrees
Therefore the angles of the triangle are:
A = 80.2 degrees, B = 55.2 degrees and C = 44.6 degrees (rounded to the tenth place).
You can draw the sketch using a compass. Draw any one of the sides and use compass on either ends of it to draw arcs corresponding to the lengths of the other two sides. The point of intersection of arcs will give you the location of the third vertices.
hope this helps.
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