Question

If the side of the triangle ABC are a=4 , b=6, c=8 , show that 4 cos B +3 cos C =2.​

Answer 1

Step-by-step explanation:

We have,

a=4,b=6,c=8

Therefore,cosB=

2aca 2 +c 2 −b 2 =

2×4×8

16+64−36 =

6444 = 16

11 and, cosC=

2ab a 2 +b 2 −c 2 = 2×4×616+36−64 =− 41

Therefore,

4cosB+3cosC=4× 16 11 − 43=2

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