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:
        






A and B are positive acute angles satisfying the equations 3cos2A + 2cos2B = 4 and (3sinA/sinB) = (2cosB/cosA), then A + 2B is equal to


A) 45


B) 60


C) 30


D) 90






8 years ago

Answers : (1)

vikas askiitian expert
509 Points
							

3cos2A + 2cos2B = 4           ................1

cos2A = 2cos2A - 1

cos2B = 2cos2B - 1

eq 1 becomes

3(1+cos2A)/2 + (1+cos2B) = 4                    .................2

 

3sinA/sinB = 2cosB/cosA

3sinAcosA =2sinB cosB                         (sin2@ = 2sin@cos@)

3sin2A = 2sin2B               .......................3

squaring eq 3

9sin22A = 4sin22B

9(1-cos22A) = 4(1-cos22B)

 9cos22A - 4cos22B = 5              ....................4

after solving 4 & 2 we get

 

cos2A = 7/9 & cos2B = 1/3

cos2A = 2cos2A-1 so

cosA = 2root2/3 , cos2B = 1/3                                  

A + 2B = cos-12root2/3 + cos-11/3                              (cos-1x+cos-1y = cos-1{xy-[(1-x2)(1-y2)]1/2}

         =cos-10 = pi/2 = 90o ...

 

8 years 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

  • 31 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