Asked by katie k
If the length of each side of a regular hexagon is 10 ft, the what is the distance from a vertex to the center?
Andrea and Carlos left the airport at the same time. Andrea flew at 180 mph on a course with bearing 80 degrees, and Carlos flew at 240 mph on a course with bearing 210 degrees. How far apart were they after 3 hr? Round the nearest 10th.
Andrea and Carlos left the airport at the same time. Andrea flew at 180 mph on a course with bearing 80 degrees, and Carlos flew at 240 mph on a course with bearing 210 degrees. How far apart were they after 3 hr? Round the nearest 10th.
Answers
Answered by
Reiny
1st one:
Too easy. Joining all vertices of a hexagon to the centre creates 6 equilateral triangles.
So if the side of the hexagon is 10 ft, all sides must be 10 ft. so the distance to the centre is 10 ft
2nd:
you have a triangle with sides 540 and 720 with an angle of 160° between them
let the distance between them be x
by cosine law:
x^2 = 540^2+720^2-2(540)(720)cos160°
= .....
you do the button-pushing.
Too easy. Joining all vertices of a hexagon to the centre creates 6 equilateral triangles.
So if the side of the hexagon is 10 ft, all sides must be 10 ft. so the distance to the centre is 10 ft
2nd:
you have a triangle with sides 540 and 720 with an angle of 160° between them
let the distance between them be x
by cosine law:
x^2 = 540^2+720^2-2(540)(720)cos160°
= .....
you do the button-pushing.
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