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18 Jul 2010, 21:29
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I couldn't find this post...could someone please explain how to arrive at the answer. Thanks!
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18 Jul 2010, 23:30
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Both OP and OQ represents the radius of the circle,
First calculate OP = sq rt ( 3+1 )
OP = OQ = 2
Now find the angle PO makes with the x-axis,
so we can say tanQ = sq rt (3) = 60 degree
this makes angles that oq makes with x-axis as 30 degree
now it is simple cos 30 = s/2
this results s = 1
simple circle & trigonometry concept
First calculate OP = sq rt ( 3+1 )
OP = OQ = 2
Now find the angle PO makes with the x-axis,
so we can say tanQ = sq rt (3) = 60 degree
this makes angles that oq makes with x-axis as 30 degree
now it is simple cos 30 = s/2
this results s = 1
simple circle & trigonometry concept
18 Jul 2010, 23:43
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Hi,
Let the angles (clockwise) at center O be u v x and y.
We have
v+x=90
u+v+x+y=180
hence u+y=90
Drop a perpendicular from P to the x axis at point T.
triangle OPT is right triangle with sides 1,sqrt(3) and 3 and hence it is a 30-60-90 triangle.
Thus angle u is 30 degrees.
Thus angle y is 60 degrees.
Now, drop a perpendicular from Q to the x at point M
Triangle OQM is right triangle with angles 30,60,90 with ratio of sides 1 sqrt(3) and 2.
Hence OM is 1 and thus s =1
regards,
Jack
Let the angles (clockwise) at center O be u v x and y.
We have
v+x=90
u+v+x+y=180
hence u+y=90
Drop a perpendicular from P to the x axis at point T.
triangle OPT is right triangle with sides 1,sqrt(3) and 3 and hence it is a 30-60-90 triangle.
Thus angle u is 30 degrees.
Thus angle y is 60 degrees.
Now, drop a perpendicular from Q to the x at point M
Triangle OQM is right triangle with angles 30,60,90 with ratio of sides 1 sqrt(3) and 2.
Hence OM is 1 and thus s =1
regards,
Jack
