if tan a=1/2 and tan b=3/8
(i)find tan(a+b)
(ii)if tan(a+b+c)=-27 find c given that c is acute.
i can do part (i) using the trig formula I get (1/2 + 3/8)/[1-(1/2)(3/8)]
I do not know how to do part (ii) though
Use the same formula for $\tan \left[(a+b) + c\right] = -27$. You will get:
$-27 = \dfrac{\tan (a+b) + \tan c}{1-\tan (a+b)\tan c}$
$-27\left[1-\tan (a+b)\tan c\right] = \tan (a+b) + \tan c$
$27\tan (a+b)\tan c - \tan c = \tan (a+b) + 27$
$\tan c = \dfrac{\tan (a+b) + 27}{27\tan (a+b) - 1}$
$c = \arctan \left( \dfrac{\tan (a+b) + 27}{27\tan (a+b) - 1} \right)$
Edit: I get $c = \dfrac{\pi}{4} = 45^\circ$