Find angle

[mathprob]

If tanA = 1/2 and tanB = 3/8:
(i) find tan(A+B)
(ii) if tan(A+B+C) = -27 find C given that C is acute.

I worked out (i) to equal 14/13

I cannot work out (ii) if anyone can help please.

 

[soroban]

Hello, mathprob!

If \(\displaystyle \tan A \,= \,\frac{1}{2}\) and \(\displaystyle \tan B \,=\,\frac{3}{8}\)

(i) find \(\displaystyle \tan(A\,+\,B)\)

(ii) if \(\displaystyle \tan(A\,+\,B\,+\,C) \:=\:-27\), find \(\displaystyle C\), given that \(\displaystyle C\) is acute.

I worked out (i) to equal \(\displaystyle \frac{14}{13}\;\;\) . . . Correct! Good work!

We use the same tangent-sum formula . . .

We are told that: \(\displaystyle \L\:\tan\bigg[(A\,+\,B)\,+\,C\bigg] \;=\;-27\)

. . . .So we have: . . . \(\displaystyle \L\frac{\tan(A+B)\,+\,\tan(C)}{1\,-\,\tan(A+B)\tan(C)} \;=\;-27\)


From (i), we have: . . . . . . . . \(\displaystyle \L\frac{\frac{14}{13}\,+\,\tan(C)}{1\,-\,\frac{14}{13}\tan(C)} \;=\;-27\)


Multiply the left side by \(\displaystyle \frac{13}{13}:\L\;\;\frac{14\,+\,13\tan(C)}{13\,-\,14\tan(C)}\;=\;-27\)

. . . . . . . . . . . . . . . . . \(\displaystyle \L14\,+\,13\tan(C) \;=\;-351\,+\,378\tan(C)\)

. . . . . . . . . . . . . . . . . . . \(\displaystyle \L-365\tan(C)\;=\;-365\)

. . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle \L\tan(C) \;=\;\frac{-365}{-365}\;=\;1\)

Therefore: \(\displaystyle \L\:C\;=\;45^o\)