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29 Jan 2009, 04:26
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Pls explain ur answers
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29 Jan 2009, 08:48
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Drop a line from P to x axis and name it m.
Also Drop a line from Q to x axis and name it n.
using 30:60:90 ratio in triangle pom:
angle pom = 30,
angle pmo = 90,
angle mpo = 60,
similarly in triangle qon,
angle qon = 60,
angle qno = 90,
angle oqn = 30,
since po = qo = 2
again using 30:60:90 ratio in triangle, on = s = 1.
on = s is opposit to angle oqn = 30.
Also Drop a line from Q to x axis and name it n.
using 30:60:90 ratio in triangle pom:
angle pom = 30,
angle pmo = 90,
angle mpo = 60,
similarly in triangle qon,
angle qon = 60,
angle qno = 90,
angle oqn = 30,
since po = qo = 2
again using 30:60:90 ratio in triangle, on = s = 1.
on = s is opposit to angle oqn = 30.
29 Jan 2009, 09:05
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Here is my approach.
OP ^2= (-sqrt(3))^2 + 1^2
-- OP =2 and OQ=2
SLOPE OF OP = Y2-Y1/X2-X1 = (1-0)/(-SQRT(3) -1) = -1/SQRT(3)
SLOPE OF OQ = SQRT(3)
becaue OP and OQ are perpendicular.
t-0/s-0 = sqrt(3) ---> t=s *sqrt(3)
s*s + t*t = 2*2
s*s (3+1)=4 --> s=1
OP ^2= (-sqrt(3))^2 + 1^2
-- OP =2 and OQ=2
SLOPE OF OP = Y2-Y1/X2-X1 = (1-0)/(-SQRT(3) -1) = -1/SQRT(3)
SLOPE OF OQ = SQRT(3)
becaue OP and OQ are perpendicular.
t-0/s-0 = sqrt(3) ---> t=s *sqrt(3)
s*s + t*t = 2*2
s*s (3+1)=4 --> s=1
