###### Please wait...
Contact Us
Contact
Need assistance? Contact us on below numbers

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

93219 24448 / 99871 78554

Mon to Sat - 10 AM to 7 PM

# Question is in the image attached

Asked by swayamagarwal2114 15th July 2022, 9:18 AM
Answered by Expert
Answer:
Let P is point of luminous source. Light ray from P incident on A is getting reflected first and
then at B is getting reflected second time and then reaches the source point P as shown in figure .

a is the distance from P to centre O and b is the distance from centre to the point
where reflected ray meets the diamenter through P .

Let us use the property of triangle as applied to ΔPQR as shown in figure.

If the line PS divides the base QR in the ration m : n and makes angle α , β and γ as shown in figure ,

then we have ,  ( m + n ) cotγ = m cotα - n cotβ

By using this property to ΔPAB as shown in left side of figure, we get

( a + b ) cotφ = (a-b) cotθ

it can be shown that φ = 90 - θ , hence cotφ = cot(90-θ) = tanθ

tan2θ = (a-b) / a+b

Answered by Expert 15th July 2022, 7:42 PM
Rate this answer
• 1
• 2
• 3
• 4
• 5
• 6
• 7
• 8
• 9
• 10

You have rated this answer /10

Your answer has been posted successfully!

### Free related questions

15th November 2021, 12:01 PM
21st November 2022, 5:29 PM
13th May 2023, 9:57 PM
RELATED STUDY RESOURCES :
JEE QUESTION ANSWERS

### Latest Questions

CBSE X Biology
Asked by Sureshgowdayd548 28th October 2020, 10:50 PM
CBSE X Biology
Asked by reenacharan2008 28th October 2020, 9:11 PM