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3 years ago

How to construct a triangle with side lengths 10 cm and 14 cm and an angle between them of 60 degrees.

Mathematics

1 answer:

Yuliya22 [10]3 years ago

4 0

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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