IIT JEE , AIEEE Forum - Mathematics , Trignometry

iit009

Q1. Find the area of the smaller portion of the disc of radius 10 cm cut off by a chord AB which subtends an angle of45/2 degree at the circumfrence.

Q2. given A+B-C=pi ,prove that sin^2A +sin^2B +sin^2C=2sinAsinBsinC

Q3. for all values of A in [0,pi/2] ,show that cos(sinA)>=sin(cosA)

Q4.The larger value of cos(log A) and log(cosA) if e^ -pi/2<A<pi/2 is.............

ashish_banga

in second it should be A+ B + C = pi

ashish_banga

3) sin X + cos X=root2 sin(pi/4+theta)

sin X + cos root 2 sinX + cos X
cos X

takin sin on both sides of equation 1 we get the required result

Please rate me if it helped u,

iit009

but in question it is given A+B-C=pi

allamraju

Q1.the ans must be

r^2/16.If we cut the circle into 16 equal parts,each subtends an angle of 45/2 at the centre

if ans is correct,pls rate me

iit009

but answer is 5(5pi-sq.root2)/2

ashish_banga

hey allam u should also substract the area of triangle
read question carefully

allamraju

Oh,sorry.I didnot read the question correctly.so,the answer is

pir^2/16-1/2r^2sin45/2(where r=10cm)

varshavallig

1) the chord subtends an angle of 45 at the centre(it is a theorem)
area of the sector=(45/180) * pie *R^2=(pie/4)*R^2
remaining area=pie*R^2-(pie/4)*R^2=(3pie/4)*R^2
area of the triangle made by the chord at the centre=
(1/2)*R^2*1/root 2
net area=(3pie/4)*R^2-R^2/2 root 2

plz correct me if m wrong
!!!!!!!!!!!!!!!!!!!

allamraju

I didnot see that the angle is at the circumference.varsha may be correct

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