
Concept explainers
To find: To Derive the formula a=b cosC+c cosB for any triangle.

Answer to Problem 60AYU
The formula is
a=b cosC+c cosB ,
Explanation of Solution
Given information:
The given triangle is ABC
Its sides and opposite angles are a, b ,c , A, B, C
Concept used:
Law of Sine:
If ABC is a triangle with sides a , b , c then
sinAa=sinBb=sinCc
The given triangle is ABC
Its sides and opposite angles are a, b ,c , A, B, C
By using the law of Sine to take the constant is k ,
sinAa=sinBb=sinCc=k
The equations are,
sinA=ak .....(1)sinB=bk .....(2)sinC=ck .....(3)
The sum of angle of the triangle ABC is,
A+B+C=180∘A=[180∘−(B+C)]
Multiplying with sine of both sides
sinA=sin[180∘−(B+C)]
The angle of sin(180∘−θ)=sinθ
sinA=sin(B+C)
By using sum to product formula,
sinA=sin(B+C)sinA=sinB.cosC+cosB.sinC
From equations ....(1), ..(2) and ..(3)
sinA=sinB.cosC+cosB.sinCak=bk.cosC+cosB.cka=b.cosC+c.cosB
Hence, the formula is a=b.cosC+c.cosB .
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Precalculus
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