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
*****************************************************************
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:
You might be interested in
Answer:
Y = 32.7°
Step-by-step explanation:
We will apply the Cosine law in the given triangle XYZ,
x = 90 ft
y = 55 ft
z = 50 ft
We have to find the measure of angle Y.
y² = x² + z² - 2xzCosY
(55)² = (90)² + (50)² - 2(90)(50)cosY
3025 = 8100 + 2500 - 9000.cosY
9000.cosY = 10600 - 3025
cosY =
= 0.84167
Y =
Y = 32.68 ≈ 32.7°
Therefore, measure of angle Y = 32.7°