WILL YOU SOLVE THIS FOR MY DAUGHTER?
TWO CHORDS (AB & AC) OF A CIRCLE WITH 5 CMS RADIUS ARE 6CMS EACH. FIND LENGTH OF BC.
TWO CHORDS (AB & AC) OF A CIRCLE WITH 5 CMS RADIUS ARE 6CMS EACH. FIND LENGTH OF BC.
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If you take one of chords, say AB, and draw 2 radii from the center to A and B, you'll have an isosceles triangle with sides OA=5, OB=5, and AB=6.
The angle between AB and OB will be 53.13°. The reason for this is that if you draw a line from the center that is perpendicular to AB, it will bisect AB and form a pair of 3, 4, 5 right triangles. In a 3,4,5 right triangle, the angle between the 3-side and the 5-side will be arcsin(4/5) = 53.13°
The angle between AB and AC will be 2 times 53.13°, or 106.26°, because AC is the same length as AB and we have a nice symmetry from this.
To find the length of chord BC, notice that we know the lengths two sides of this triangle (both 6) and the included angle (106.26°), We can therefore use the law of cosines to determine the length of the remaining side.
The general form of the law of cosines is c² = a² + b² - 2ab(cos C) where C is the angle between sides a & b, and opposite of side c.
Plugging our values into this formula.....
c² = 6² + 6² - 2(6)(6)(cos 106.26°)
c² = 6² + 6² - 2(6)(6)(-0.28)
c² = 36 + 36 - 72(-0.28)
c² = 72 + 72(0.28)
c² = 72 + 72(0.28)
c² = 72 + 20.16
c² = 92.16
c = 9.6
ANSWER
Length of BC = 9.6 cm
Hope all that made sense!