Let A, B, and C be the angles of a triangle. Given that tan (A/2) = 1/4 and tan (B/2) = 1/5, find tan (C/2).
Using the half-angle formulae:
tan(A)=2tan(A2)1−tan2(A2)=2(14)1−(14)2=815 .........(1)
tan(B)=2tan(B2)1−tan2(B2)=2(15)21−(15)2=512 ..........(2)
tan(C)=2tan(C2)1−tan2(C2) ........(*)
Since the angles in a triangle sum to 180 degrees, then:
A+B+C=180⟺tan(A+B+C)=0 (Take tan of both sides).
Thus, consider: tan((A+B)+C)=tan(A+B)+tan(C)1−tan(A+B)tan(C)=0 ........(**)
Now consider tan(A+B), and using (1) and (2) gives:
tan(A+B)=tan(A)+tan(B1−tan(A)tan(B)=815+5121−(815∗512)=171140 ......(***)
Thus, substituting (***) in (**) gives:
171140+tan(C)=0⟺tan(C)=−171140
Next, consider (*) and let t=tan(C2)
Then,
−171140=2t1−t2⟺171t2−171=280t
Then, 171t2−280t−171=0⟺(19t+9)(9t−19)=0
Thus, t1=199,t2=−919
But, C is a positive angle (Angle in a triangle), hence t=tan(C2)>0
Therefore, only t1=199 is the desired solution.
Thus, tan(C2)=199
Hope this helps, and I think there is a simpler solution than this!
Using the half-angle formulae:
tan(A)=2tan(A2)1−tan2(A2)=2(14)1−(14)2=815 .........(1)
tan(B)=2tan(B2)1−tan2(B2)=2(15)21−(15)2=512 ..........(2)
tan(C)=2tan(C2)1−tan2(C2) ........(*)
Since the angles in a triangle sum to 180 degrees, then:
A+B+C=180⟺tan(A+B+C)=0 (Take tan of both sides).
Thus, consider: tan((A+B)+C)=tan(A+B)+tan(C)1−tan(A+B)tan(C)=0 ........(**)
Now consider tan(A+B), and using (1) and (2) gives:
tan(A+B)=tan(A)+tan(B1−tan(A)tan(B)=815+5121−(815∗512)=171140 ......(***)
Thus, substituting (***) in (**) gives:
171140+tan(C)=0⟺tan(C)=−171140
Next, consider (*) and let t=tan(C2)
Then,
−171140=2t1−t2⟺171t2−171=280t
Then, 171t2−280t−171=0⟺(19t+9)(9t−19)=0
Thus, t1=199,t2=−919
But, C is a positive angle (Angle in a triangle), hence t=tan(C2)>0
Therefore, only t1=199 is the desired solution.
Thus, tan(C2)=199
Hope this helps, and I think there is a simpler solution than this!