Construct a triangle ABC in which AB = 5 cm and BC = 4.6 cm and AC = 3.7 cm
Answers
In order to construct the diameter of the circumscribed circle we construct a right triangle with angles 30 and 60 degrees and the leg across 60 degrees equal to 3.7 cm. This is an easy construction problem. We have constructed a line segment with the length 2R. It id the hypotenuse of the triangle. This is all we need. You can now forget about the rest of the triangle.
We now construct a circle with diameter 2R and divide the circle into three equal parts. Mark two of the three division points A and B. Connect A and B. AB =3.5, because law of sine in valid. What is left is to construct point C. It is somewhere on the circumference on the larger arc AB. Draw a perpendicular to the line AB and then a line parallel to AB such that its distance from AB is equal to the given height.
There are three cases.
The line is outside the circle, then there is no solution.
The line is tangent to the circle. Then this tangent point is C, one solution.
The line crosses the circle in two points. Then there are two possible positions for point C, two solutions.
Answer:
In order to construct the diameter of the circumscribed circle we construct a right triangle with angles 30 and 60 degrees and the leg across 60 degrees equal to 3.7 cm. This is an easy construction problem. We have constructed a line segment with the length 2R. It id the hypotenuse of the triangle. This is all we need. You can now forget about the rest of the triangle.
We now construct a circle with diameter 2R and divide the circle into three equal parts. Mark two of the three division points A and B. Connect A and B. AB =3.5, because law of sine in valid. What is left is to construct point C. It is somewhere on the circumference on the larger arc AB. Draw a perpendicular to the line AB and then a line parallel to AB such that its distance from AB is equal to the given height.
There are three cases.
The line is outside the circle, then there is no solution.
The line is tangent to the circle. Then this tangent point is C, one solution.
The line crosses the circle in two points.
Step-by-step explanation: