How do you show that for triangle ABC, $\frac{a \pm b}{c} = \frac{sin A \pm sin B}{sin C}$ $(a+-b)/c=(sinA+-sinB)/sinC$ ?

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How do you show that for triangle ABC, $\frac{a \pm b}{c} = \frac{sin A \pm sin B}{sin C}$ $(a+-b)/c=(sinA+-sinB)/sinC$ ?

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Answer 1 · 2016-10-21T12:41:49

See below.

Explanation:

If $λ = \frac{n_{1}}{n_{2}} = \frac{m_{1}}{m_{2}}$ $lambda=n_1/n_2=m_1/m_2$ then $λ = \frac{n_{1} \pm m_{1}}{n_{2} \pm m_{2}}$ $lambda = (n_1pmm_1)/(n_2pmm_2)$ because substituting

$\frac{λ n_{2} \pm λ m_{2}}{n_{2} \pm n_{2}} = λ$ $(lambda n_2pmlambdam_2)/(n_2pmn_2)= lambda$
Now considering true the sinus law

$\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$ $sinA/a=sinB/b=sinC/c$ and arranging we have

$\frac{sin C}{c} = \frac{sin A \pm sin B}{a \pm b}$ $sinC/c=(sinApmsinB)/(apmb)$ following

$\frac{a \pm b}{c} = \frac{sin A \pm sin B}{sin C}$ $(a+-b)/c=(sinA+-sinB)/sinC$

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How do you show that for triangle ABC, $\frac{a \pm b}{c} = \frac{sin A \pm sin B}{sin C}$ $(a+-b)/c=(sinA+-sinB)/sinC$ ?

1 Answer

Explanation:

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How do you show that for triangle ABC, a±bc=sinA±sinBsinC(a+-b)/c=(sinA+-sinB)/sinC?

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How do you show that for triangle ABC, $\frac{a \pm b}{c} = \frac{sin A \pm sin B}{sin C}$ $(a+-b)/c=(sinA+-sinB)/sinC$ ?