CLEAR
Toprankers-Info

+91 7676564400

Mobile App

Get it on Google Play
Discussions
Select Date
Tags:

Articles

Discussions
Submit Discussion
Deepak

· started a discussion

· 1 Months ago

can anyone prove it wrong plz let me know, plz reply to it (Question: 269935.0 )

1/2sinq.xx = 1/2.x/2.root3x
sinq=sin60
q=60
center L 60.
PLZ DO REPLY

Question:

Find the value of the angle subtended by the chord at the centre of the circle, if the length of chord is  3   x and radius is x.

Options:
A)

60º

B)

30º

C)

120º

D)

90º

Solution:

Ans: (c)

  so,  angle on center=120º

Knowledge Expert

· commented

· 1 Months ago

Dear Student
With the help pf pythagorus theorum we can find the value of OB^2={(AO)^2-(AB)^2}
OB^2={X^2-(root3/x)^2}
OB=X/2
NOW, considering the triangle AOB
Taking angle(AOB) as "Theta"
Cos(angleAOB)=BO/AO=(X/2)/(X)=1/2
Cos(AngleAOB)=1/2
AngleAOB=60 degrees
The angle subtended by the chord on the centre of the circle(O) is AngleAOC
AngleAOC=2 X AngleAOB
AngleAOC=2 X 60degrees
AngleAOC=120 degrees
Keep solving, Keep learning
Best wishes
Regards
Team TopRankers
-logo
We at aim to provide most comprehensive content & test for practice which is carefully divided into various chapters and topics to help you focus on your weak areas.Our Practice mock test gives you unmatched analytics to help you realize your strong areas,time management and improvement areas.Our Short & Crisp notes on each topic will help you to quickly revise the topic along with the tips & tricks of the exams.
More About
-Info
-Info

Mobile App

Get it on Google Play
About Us Campus Ambassador Careers Contact Us Refund Policy

All Rights Reserved Top Rankers

User Policy
Disclaimer
Terms & Conditions