 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

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
```
In any triangle ABC, prove that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.
In any triangle ABC, prove that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.

```
4 months ago Anand Kumar Pandey
3210 Points
```							Welcome to AskIITiansIn any triangle ABC,a/sin A = b/sin B = c/sin C = ka = k sin A, b = k sin B, c = k sin CLHS= a sin (B – C) + b sin (C – A) + c sin (A – B)= k sin A [sin B cos C – cos B sin C] + k sin B [sin C cos A – cos C sin A] + k sin C [sin A cos B – cos A sin B]= k sin A sin B cos C – k sin A cos B sin C + k sin B sin C cos A – k sin B cos C sin A + k sin C sin A cos B – k sin C cos A sin B= 0= RHSHence proved that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.Thanks
```
4 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Analytical Geometry

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

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