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

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Answer 1 · 2016-10-21T12:41:49

See below.

If $lambda=n_1/n_2=m_1/m_2$ then $lambda = (n_1pmm_1)/(n_2pmm_2)$ because substituting

$(lambda n_2pmlambdam_2)/(n_2pmn_2)= lambda$
Now considering true the sinus law

$sinA/a=sinB/b=sinC/c$ and arranging we have

$sinC/c=(sinApmsinB)/(apmb)$ following

$(a+-b)/c=(sinA+-sinB)/sinC$

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