MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 11
        
If the line x=y=z intersect the line sinA x+sinB y+sinC z=2d^2,sin2A x+sin2B y+sin2C z=d^2 then sinA/2sinB/2sinC/2 is equal to where (A+B+C=pi)
 
10 months ago

Answers : (1)

Arun
22540 Points
							
i) Let x = y = z = k -------- (1) 

ii) Substituting in the given other two equations from (1) and reducing, we get: 
sin(A) + sin(B) + sin(C) = 2d²/k ------ (2) 

and sin(2A) + sin(2B) + sin(2C) = d²/k -------- (3) 

iii) Dividing (3) by (2): 
[sin(2A) + sin(2B) + sin(2C)]/[sin(A) + sin(B) + sin(C)] = 1/2 ---------- (4) 

iv) You may be aware from trigonometric identities for the given condition A + B + C = π, that 
sin(2A) + sin(2B) + sin(2C) = 4*sin(A)*sin(B)*sin(C) and 
sin(A) + sin(B) + sin(C) = 4*cos(A/2)*cos(B/2)*cos(C/2) 

Substituting these in (4) above, 
4*sin(A)*sin(B)*sin(C)/ 4*cos(A/2)*cos(B/2)*cos(C/2) = 1/2 

==> sin(A)*sin(B)*sin(C)/cos(A/2)*cos(B/2)*c... = 1/2 

==> [2sin(A/2)cos(A/2)*2sin(B/2)cos(B/2)*2si... = 1/2 
{Application of multiple angle identities} 

On cancellation of the corresponding terms and simplification we get, 
sin(A/2)*sin(B/2)*sin(C/2) = 1/16
10 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 19 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details