Question

Find measure of side AB round to nearest tenth

Answer 1

Use:

$|AB|^2=|AC|^2+|BC|^2-2|AC||BC|\cos(m\angle ACB)$

We have:

$|AC|=13;\ |BC|=29;\ m\angle ACB=41^o$

$\cos41^o\approx0.7547$

substitute:

$|AB|^2=13^2+29^6-2\cdot13\cdot29\cdot0.7547\\\\|AB|^2=440.9562\\\\|AB|\approx21$