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P and Q be centers of two circle having radius 200cms. These circle intersect each other at some point A and B. Length of PQ is 250 cms. What will the angle AQP be?
a) between 0 to 45
b) between 0 to 30
c) between 0 to 60
d) between 0 to 75
Read Solution (Total 8)
-
- draw the diagram, in triangle AQP, AP = 200 , AQ = 200 and PQ = 250
cos(AQP)= (200^2+250^2-200^2)/(2*200*250)= 5/8 = .625 [using cal, cos(AQP)= 51]
or,
as cos45=.7 & cos60=.5 so, cos(AQP)=.625 lies betn angles 45 & 60
c) between 0 to 60
- 10 years agoHelpfull: Yes(29) No(6)
- ANS IS a)between 0 to 45
EXPLANATION:
since if we draw the figure, then AP= 200 , AQ=200 and PQ=250
triangle property of 45- 45 - 90 is that H=L * root(2)
H= hypo and L= Leg.therefore H/L= root(2)
here H=250 and L=200 and H/L= 250/200= 1.25
and the value of root(2)== 1.44
so H/L= approx root(2).
Therefore the angle between H and L = near about 44 degree
so the ans will be 0 to 45.
- 10 years agoHelpfull: Yes(10) No(12)
- If you draw the diagram,A is the intersecting point, so if we join centres of both circles to it, it will form a 90 degree angle there(PAQ),(as a tangent and radius are always perpendicular to each other)
So,
In triangle PAQ,
PA=AQ=200;
PQ=250;
Then,
If we consider angle AQP as 'x';
sin x= opp/hypo = 200/250 = 0.8;
therefore, x = 53.13 degrees.
So the answer is c, between 0 to 60 - 10 years agoHelpfull: Yes(8) No(0)
- a is answer
- 10 years agoHelpfull: Yes(3) No(4)
- d will be the answer
- 10 years agoHelpfull: Yes(2) No(3)
- c will be the answer 100%
- 10 years agoHelpfull: Yes(2) No(1)
- as two of the sides are same so the angles opposite to these sides should be same.angle PAQ is 90 deg,so angle aqp surely be 45 deg.so the correct ans is option a.if anyone thinks m wrong then plz xplain
- 10 years agoHelpfull: Yes(1) No(0)
- First find length of common chord 2*r1*r2/d (d= distance between two centres)
then sin (x)=160/200
x=53.13 - 9 years agoHelpfull: Yes(0) No(0)
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