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# geometry

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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).

Apr 28, 2022

#1
+1

Using the half-angle formulae:

$$tan(A)=\frac{2tan(\frac{A}{2})}{1-tan^2(\frac{A}{2})}=\frac{2(\frac{1}{4})}{1-(\frac{1}{4})^2}=\frac{8}{15}$$  .........(1)

$$tan(B)=\frac{2tan(\frac{B}{2})}{1-tan^2(\frac{B}{2})}=\frac{2(\frac{1}{5})^2}{1-(\frac{1}{5})^2}=\frac{5}{12}$$ ..........(2)

$$tan(C)=\frac{2tan(\frac{C}{2})}{1-tan^2(\frac{C}{2})}$$                ........(*)

Since the angles in a triangle sum to 180 degrees, then:

$$A+B+C=180 \iff tan(A+B+C)=0$$  (Take tan of both sides).

Thus, consider:                        $$tan((A+B)+C)=\frac{tan(A+B)+tan(C)}{1-tan(A+B)tan(C)}=0$$        ........(**)

Now consider $$tan(A+B)$$, and using (1) and (2) gives:

$$tan(A+B)=\frac{tan(A)+tan(B}{1-tan(A)tan(B)}=\frac{\frac{8}{15}+\frac{5}{12}}{1-(\frac{8}{15}*\frac{5}{12})}=\frac{171}{140}$$      ......(***)

Thus, substituting (***) in (**) gives:

$$\frac{171}{140}+tan(C)=0 \iff tan(C)=-\frac{171}{140}$$

Next, consider (*) and let  $$t=tan(\frac{C}{2})$$

Then,

$$-\frac{171}{140}=\frac{2t}{1-t^2} \iff 171t^2-171=280t$$

Then,                              $$171t^2-280t-171=0 \iff(19t+9)(9t-19)=0$$

Thus,                                                       $$t_1=\frac{19}{9},t_2=-\frac{9}{19}$$

But, C is a positive angle (Angle in a triangle), hence $$t=tan(\frac{C}{2})>0$$

Therefore, only $$t_1=\frac{19}{9}$$ is the desired solution.

Thus,   $$tan(\frac{C}{2})=\frac{19}{9}$$

Hope this helps, and I think there is a simpler solution than this!

Apr 29, 2022

#1
+1

Using the half-angle formulae:

$$tan(A)=\frac{2tan(\frac{A}{2})}{1-tan^2(\frac{A}{2})}=\frac{2(\frac{1}{4})}{1-(\frac{1}{4})^2}=\frac{8}{15}$$  .........(1)

$$tan(B)=\frac{2tan(\frac{B}{2})}{1-tan^2(\frac{B}{2})}=\frac{2(\frac{1}{5})^2}{1-(\frac{1}{5})^2}=\frac{5}{12}$$ ..........(2)

$$tan(C)=\frac{2tan(\frac{C}{2})}{1-tan^2(\frac{C}{2})}$$                ........(*)

Since the angles in a triangle sum to 180 degrees, then:

$$A+B+C=180 \iff tan(A+B+C)=0$$  (Take tan of both sides).

Thus, consider:                        $$tan((A+B)+C)=\frac{tan(A+B)+tan(C)}{1-tan(A+B)tan(C)}=0$$        ........(**)

Now consider $$tan(A+B)$$, and using (1) and (2) gives:

$$tan(A+B)=\frac{tan(A)+tan(B}{1-tan(A)tan(B)}=\frac{\frac{8}{15}+\frac{5}{12}}{1-(\frac{8}{15}*\frac{5}{12})}=\frac{171}{140}$$      ......(***)

Thus, substituting (***) in (**) gives:

$$\frac{171}{140}+tan(C)=0 \iff tan(C)=-\frac{171}{140}$$

Next, consider (*) and let  $$t=tan(\frac{C}{2})$$

Then,

$$-\frac{171}{140}=\frac{2t}{1-t^2} \iff 171t^2-171=280t$$

Then,                              $$171t^2-280t-171=0 \iff(19t+9)(9t-19)=0$$

Thus,                                                       $$t_1=\frac{19}{9},t_2=-\frac{9}{19}$$

But, C is a positive angle (Angle in a triangle), hence $$t=tan(\frac{C}{2})>0$$

Therefore, only $$t_1=\frac{19}{9}$$ is the desired solution.

Thus,   $$tan(\frac{C}{2})=\frac{19}{9}$$

Hope this helps, and I think there is a simpler solution than this!

Guest Apr 29, 2022
#2
+124701
0

Very nice solution to  a very tricky one  !!!!!

CPhill  Apr 29, 2022
#3
0

Thanks Cphill!

Guest Apr 29, 2022