Question

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

Answer 1

See below.

Explanation:

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`