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
`        If ABC.be triangle then acosBcosC+bcosCcosA+ccosAcosB=abc/4R^2 prove it `
11 months ago

20 Points
```							 =cosAcosBcosC[a/cosA + b/cosB + c/cosC] using Sine rule:=cosAcosBcosC(2R)[tanA + tanB + tanC]{R= circumradius }{tanA + tanB + tanC = tanAtanBtanC}=cosAcosBcosC(2R)sinAsinBsinC/cosAcosBcosC=2RsinAsinBsinC again using the sine rule= 2Rabc/2R*2R*2R = abc/4R2  where a,b,c = sides of triangle
```
one month 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