Question

Prove that in any triangle ABC,sinA/a=sinB/b=sinC/c

Answer 1

Consider a triangle ABC of sides $\sf{a,\ b,\ c}$ opposite to the angles $\sf{A,\ B,\ C}$ respectively, with altitudes $\sf{h_a,\ h_b,\ h_c}$ drawn from respective angles to the opposite sides. Then,

$\sf{\sin A=\dfrac {h_b}{c}}\\\\\\\sf{\sin B=\dfrac {h_c}{a}}\\\\\\\sf{\sin C=\dfrac {h_a}{b}}$

Then the area of the triangle ABC can be any among,

$\sf{Area=\dfrac {1}{2}\cdot a\cdot h_a}\\\\\\\sf{Area=\dfrac {1}{2}\cdot b\cdot h_b}\\\\\\\sf{Area=\dfrac {1}{2}\cdot c\cdot h_c}$

Then we have the following,

$\sf{b\cdot h_b=c\cdot h_c=a\cdot h_a}$

Dividing each by $\sf{abc,}$

$\sf{\dfrac {b\cdot h_b}{abc}=\dfrac {c\cdot h_c}{abc}=\dfrac {a\cdot h_a}{abc}}\\\\\\\sf{\dfrac {h_b}{ac}=\dfrac {h_c}{ab}=\dfrac {h_a}{bc}}\\\\\\\sf{\dfrac {h_b}{c}\cdot\dfrac{1}{a}=\dfrac {h_c}{a}\cdot\dfrac{1}{b}=\dfrac {h_a}{b}\cdot\dfrac{1}{c}}\\\\\\\sf{\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}}$

Hence Proved!