In a triangle $\triangle ABC$ it is given that $(\sin A+\sin B):(\sin B+\sin C):(\sin C+\sin A)=9:10:11$. Find the value of $480\cos A$
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Tags: trigonometry
Farenhajt
03.06.2014 11:16
Using the Sine Law, this turns into $(2s-c):(2s-a):(2s-b)=9:10:11$, hence $2s-c=9k, 2s-a=10k,2s-b=11k$. Adding them up, we get $4s=30k\iff 2s=15k$, thus $a=5k\land b=4k\land c=6k$.
Hence $480\cos A=480\cdot{16+36-25\over 2\cdot 5\cdot 4}=324$
SMOJ
03.06.2014 11:20
Farenhajt wrote: Hence $480\cos A=480\cdot{16+36-25\over 2\cdot 5\cdot 4}=324$ You mean $2\times 6\times 4$?
Farenhajt
03.06.2014 12:06
Indeed Hence the result is $270$. Thanks
Krishijivi
14.10.2023 16:44
( sin A+ sin B)/ ( sin B+ sin C)= 9/10 10 sin A= 9 sin C - sin B....(1) sin C+ sin B)/ ( sin A+ sin C)= 10/11 10 sin A= 11 sin B+ sin C Subtracting the equations, 2 sin C= 3 sin B sin C/ sin B= 3/2 c/b=3/2, c= 3b/2 sin A+ sin B)/ ( sin A+ sin C)= 9/11 2sin A+ 11 sin B=9 sin C....(3) (1)-(3) 8 sin A= 10 sin B sin A/ sin B=5/4 a/b=5/4 a=5b/4 cos A=(b²+c²-a²)/2bc = {b²+(9b²/4)-(25b²/16)}/ 2b*(3b/2) =(27b²)/ 48 b² =9/16 480 cos A= 480*(9/16) = 270 @Krishijivi