 Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

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
`        Find the answer fastly and get the correct answer  immediately `
one year ago

```							 a cos A + b cos B + c cos C, ... where ... a/sin A = ... = 2R= R [ 2 sin A cos A + 2 sin B cos B + 2 sin C cos C ]= R [ ( sin 2A + sin 2B ) + sin 2C ]= R [ 2 sin (A+B)· cos(A-B) + sin 2C ]= R [ 2 sin C. cos(A-B) + 2 sin C cos C ]= R sin C [ cos(A-B) + cos C ] ... here .. cos C = cos [ π - (A+B) ] = - cos (A+B)= R sin C [ cos(A-B) - cos(A+B) ]= R sin C [ 2 sin A sin B ]= 2 ( 2R sin A ) sin B sin C= 2 a sin B sin C=2a [ 2Δ / ca ] [ 2Δ / ab ] ...... from (2)= 8 Δ² / ( abc )= r/R(a+b+c) Hence a cos A + b cosB + c cosC / 2s = r/R
```
one year ago Deepak Kumar Shringi
4397 Points
```							https://www.askiitians.com/forums/Trigonometry/in-a-triangle-abc-the-value-of-acosa-bcosb-ccosc-a_109066.htmplease go through this
```
one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Trigonometry

View all Questions »  ### 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