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

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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step-by-step explanation:

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step-by-step explanation:

u take $30 minus $24. the answer/original($30)