5
$\begingroup$

A chord of a circle subtends an angle of 89 degrees at its centre. Find the length of the chord given that the circle's diameter is 11.4 cm.

The problem I have here is that I can't visualise this question. I've tried drawing it but gets all messy and I get confused so can someone please draw the diagram for me? That's all I need because I can work from there on after.

$\endgroup$
  • $\begingroup$ This may be of interest (even if not for this particular question). ${}\qquad{}$ $\endgroup$ – Michael Hardy Feb 1 '15 at 3:56
3
$\begingroup$

This is how I interpret the problem:

sketch of the problem

The task is to find the length of the red line.

$\endgroup$
1
$\begingroup$

There is a formula for that,

$$ 2 \cdot 5.7 \cdot \sin \left(\frac{89 \pi }{360} \right) = 7.9903656130183 $$ enter image description here

Where A is the angle in radians and r is the radius.

$\endgroup$
  • $\begingroup$ How can we derive the above formula? $\endgroup$ – Gauz Mar 14 '17 at 19:28
  • $\begingroup$ @Gauz it looks like the law of sines $\endgroup$ – bobbym Mar 14 '17 at 22:11
  • $\begingroup$ Not sure what law of sines is but if you drop a perpendicular from the center of the circle to the chord, property says that the perpendicular bisects the chord. So we can consider one of the triangle, and find out half of the chord length which is $$sin(A/2) = l/r$$ Rearranging for l and multiplying by 2, we get the full chord length. $\endgroup$ – Gauz Mar 15 '17 at 17:46
  • $\begingroup$ Hi Gauz; en.wikipedia.org/wiki/Law_of_sines $\endgroup$ – bobbym Mar 15 '17 at 21:32
0
$\begingroup$

AB is the chord.Angle AOB = 89 degrees

The way to solve it is : Use the Formula: 2r sin A/C

$OA = OB = 5.7$cm as it is the radius. So, you would get:

$2 \cdot 5.7 \cdot \sin \left(\frac{89 \pi }{360} \right) = 7.9903656130183$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.