21-05-10 : QOD MATHS

Not QoD. sibal's doubts.

at least try, i will post the solution today

lol sibal...you're acting like you've been granted the permission to post QoDs...
But you can ask nishant bhaiya for that if you're so desperate :P

a- r, b- s, c- q, d- p

EMF is correct....

In such cases, the required angle α can be found easily by means of the graph of the function y = sin x. Plot the number 10 on x-axis at the point P and find sin 10 geometrically. (It is the ordinate PN) and then draw the horizontal line (i.e. the line y = sin 10). The abscissa of one of the points intersection of this straight line with the graph lies at Q in the interval [–/2, /2] and its sine is equal to sine 10. The abscissa of Q, say , is the desired angle. It can be found easily by geometrical reasoning. It is easy to see that the point α and 10 are symmetric about the point 3/2 so that 10 – 3/2 = 3/2 – , whence
α=3π-10

Try to solve other parts