AB is a diameter of a circle with centre at O. DC is a chord such that DC is parallel to AB. If angle BAC=20, then angle COD is
@ravi.theja said:.(theta)/360*pi*(1)^2 ==> dis one is formula for area of sector na....here it is nt askd so jus finding s1 is enough
yaar s1 to area hua na...we have to find the angle made by this sector...so theta to nikalna hi hoga na??
@mailtoankit said:yaar s1 to area hua na...we have to find the angle made by this sector...so theta to nikalna hi hoga na??
Fine Fine 😃 understood !! u r rite.. mind sleeping 
@mailtoankit said:kaise kiya bhai...yahan to mein ghoom gaya hoon angles mein..
CAB 20 hai to arc CB is 40. to arc AB hua 40. ab arc BC=180-40-40=100
@audiq7 said:CAB 20 hai to arc CB is 40. to arc AB hua 40. ab arc BC=180-40-40=100
kya yaar..itna simple tha.
..saala pata nahi kya-kya angle bana ke dekh raha tha.
..thanks bhai.
..saala pata nahi kya-kya angle bana ke dekh raha tha...thanks bhai.
From a circular sheet of paper of radius 10 cm , a sector of area 40 % is removed. if the remaining part is used to make a conical surface then ratio of the radius and height of the cone is
@ravi.theja
Shld be 3:4
Now, Radius of the paper = Slant height of the Cone(l)
Now, Circumference of the remaining part of the sheet = Circumference of the base of the cone..
=> 2pi*60/100*r = 2pi*r'[Here, r=radius of the circle, r'= radius of the cone]
Thus, r' = 3/5*r
Now, Slant Height^2 = Radius of the cone^2 + Height of the cone^2
=> r^2 = 9/25*r^2 + h^2
=>h = 4/5*r
Thus, Required Ratio = r'/h = (3/5*r)/(4/5*r) = 3:4..
@ravi.theja said:From a circular sheet of paper of radius 10 cm , a sector of area 40 % is removed. if the remaining part is used to make a conical surface then ratio of the radius and height of the cone is
3:4?
length of the arc = (1/2)*l*r = 40pi---> l = 8pi
circumference of remaining part = 20pi - 8pi = 12pi
radius of cone = 2*pi*r = 12pi----> r = 6
slant height = 10
10^2 = h^2 + 6^2-----> h = 8
r/h = 6/8 = 3/4
PQ is a chord of 6cm of a circle of radius of 5cm.Tangents to circle at P and Q meet at T..find lenght of TP
@TootaHuaDil said:If binary nos from 100 to 1000000 are written find the no of 1s in it.
No. of times a digit occurs in 'n' digit number in base b=b^(n-1)+(n-1)*(b-1)*b^(n-2)
here b=2
n=3,4,5,6
So ans=8+20+48+112+1=189
@bs0409 said:No. of times a digit occurs in 'n' digit number in base b=b^(n-1)+(n-1)*(b-1)*b^(n-2)here b=2n=3,4,5,6So ans=8+20+48+112+1=189
Please explain the bold part :O
@ravi.theja said:PQ is a chord of 6cm of a circle of radius of 5cm.Tangents to circle at P and Q meet at T..find lenght of TP
TP = 15/4?
@bs0409 said:No. of times a digit occurs in 'n' digit number in base b=b^(n-1)+(n-1)*(b-1)*b^(n-2)here b=2n=3,4,5,6So ans=8+20+48+112+1=189
kya formula hai...
*take a bow*
@TootaHuaDil said:Please explain the bold part
Here we have 3 digit numbers i.e. 100 to 111
4 digit numbers i.e. 1000 to 1111
5 digit numbers i.e. 10000 to 11111
6 digit numbers i.e. 100000 to 111111
and 1 seven digit number i.e. 1000000