In a rhomboid the obtuse angle is 123 deg, the longer side is 50mm, and the longer diagonal is 80mm?

Question

Find the shorter side of the rhomboid.

Answer 1

break the rhomboid into two triangles, by drawing a line along the longer diagonal

You now can solve for the short side by the cosine law

c^2 = a^2 + b^2 - 2ab*cosC
80^2 = 50^2 + b^2 - 2(50)(b)*cos(123)
6400 - 2500 = b^2 +54.46b
b^2 + 54.46b -3900 = 0

From quadratic eq'n
b = 40.898 or -95.358 (inadmissable)

b = 40.9 mm
short side of rhomboid = 40.9 mm

Answer 2

You can use the law of cosines, because you know the lengths of two sides and one angle of a triangle. If you draw in the diagonal of length 80, you create a triangle where you know the length of another side is 50. You also know the measure of one angle, because adjacent angles in a rhomboid are supplementary; if the obtuse angle is 123, the acute angle is 180 - 123 = 67. So you have known side 50 and unknown side x with angle 67 between them and known side 80 opposite that angle.

The law of cosines states that c^2 = a^2 + b^2 - 2ab*cos(C), where C is the angle opposite side c and a and b are the other two sides. Here we have C = 67, c = 80, a = 50, b = x. That gives us 80^2 = 50^2 + x^2 - 2*50x*cos(67) ==> 6400 = 2500 + x^2 + 100x*cos(67) ==> x^2 + 39.7x - 3900 = 0, a quadratic equation that you must solve. You'll want to use the quadratic formula, x = [-b +/- sqrt(b^2 - 4ac)] / 2a, where a = 1, b = 39.7, and c = -3900. That gives us x = [-39.7 + sqrt(39.7^2 - 4*(-3900))] / 2 = (-39.7 + sqrt(17127)) / 2 = 45.9 mm. Note that I knew I would use + instead of -, because I knew my final answer had to be positive.