How to construct a triangle with side lengths 10 cm and 14 cm and an angle between them of 60 degrees.
1 answer:
Answer:
We draw line AB which is perpendicular to the 14 cm side
Since Angle C is 60 degrees that makes angle CAB = 30 degrees
Triangle CAB is a 30 60 90 triangle so line CB is half the hypotenuse or 5 cm
Line BD equals 9 cm
Line AB^2 = 10^2 - 5^2 = 75
Line AD^2 = AB^2 + BD^2
Line AD^2 = 75 + 81
Line AD^2 = 156
Line AD = 12.4899959968
Line AB = Sqr root (75) = 8.6602540378
Angle D = arc sine (8.6602540378 / 12.4899959968)
Angle D = 43.898 degrees
Angle A = 180 - 60 - 43.898 = 76.102 degrees
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Actually this could have been done a little easier by using the Law of Cosines and then the Law of Sines, but I just thought I'd show another way to solve this.
Step-by-step explanation:
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