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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Answer18:
The quadrilateral ABCD is not a parallelogram
Answer19:
The quadrilateral ABCD is a parallelogram
Step-by-step explanation:
For question 18:
Given that vertices of a quadrilateral are A(-4,-1), B(-4,6), C(2,6) and D(2,-4)
The slope of a line is given m=
Now,
The slope of a line AB:
m=
m=
m=
The slope is 90 degree
The slope of a line BC:
m=
m=
m=
The slope is zero degree
The slope of a line CD:
m=
m=
m=
The slope is 90 degree
The slope of a line DA:
m=
m=
m=
m=
The slope of the only line AB and CD are the same.
Thus, The quadrilateral ABCD is not a parallelogram
For question 19:
Given that vertices of a quadrilateral are A(-2,3), B(3,2), C(2,-1) and D(-3,0)
The slope of a line is given m=
Now,
The slope of a line AB:
m=
m=
m=
The slope of a line BC:
m=
m=
m=
m=3
The slope of a line CD:
m=
m=
m=
The slope of a line DA:
m=
m=
m=3
The slope of the line AB and CD are the same
The slope of the line BC and DA are the same
Thus, The quadrilateral ABCD is a parallelogram