Law of Tangent

[nycmathguy]

In trigonometry, we are surely to come across the law of sines and law of cosines. However, there is a law of tangent that is never taught in trigonometry.

1. Why is the law of tangents ignored?

2. Solve the following using the law of tangents.

In triangle ABC, little a = 321, little b = 234 and angle C = 71°. Find angle A, angle B and little c to the nearest tenth.

Note: It is called the Law of Tangents not the Law of Tangent. I forgot the letter "s" in the title for this thread.

 

[MathLover1]

In triangle ABC, with sides a, b, and c opposite the respective angles A, B, and C, the law of tangents states:

(a-b)/(a+b)=tan((A-B)/2)/tan((A+B)/2)….(1)

Similarly for other sides,

(b-c)/(b+c)=tan((B-C)/2)/tan((B+C)/2)….(2)

(c-a)/(c+a)=tan((C-A)/2)/tan((C+A)/2) …..(3)


if a = 321, b = 234 and angle C = 71°

(234-c)/(234+c)=tan((B-71)/2)/tan((B+71)/2)
c = 234 sin(71) (1/sin(B))...........(2a)

(c-321)/(c+321)=tan((71-A)/2)/tan((71+A)/2)
c = 321 sin(71) (1/sin(A))................(3a)

from (2a) and (3a) we have

234 sin(71) (1/sin(B))=321 sin(71) (1/sin(A))..........simplify

234/sin(B)=321/sin(A)

234sin(A)=321sin(B)

sin(A)=(321/234)sin(B)
sin(A)=(107/78)sin(B)
A=sin^-1((107/78)sin(B))

go to
c = 321 sin(71) (1/sin(A))................(3a), substitute A
c = 321 sin(71) (1/sin(sin^-1((107/78)sin(B))))
c = 321sin(71)(107/78) sin(B)..........(3b)

from (2a) and (3b) we have

234 sin(71) (1/sin(B))=321sin(71)(107/78) sin(B).......solve for B

234/sin(B)=321(107/78) sin(B)
sin^2(B)=234/(321(107/78))
sin^2(B)=6084/11449

sin(B)=sqrt(6084/11449)

sin(B)=0.7289719626168224

B=sin^-1(0.7289719626168224)
B=46.80027915261779°

since given C = 71°, then angle A will be

A=180-( 71°+46.80027915261779°)
A=180°-( 117.80027915261779° )
A=62.19972084738221°

go to
c = 321 sin(71) (1/sin(A)), substitute A
c = 321 sin(71) (1/sin(62.19972084738221))
c = 343.11414209938454

angle A, angle B and little c to the nearest tenth.
A=62.2°
B=46.8°
c = 343.1

 

[nycmathguy]

View attachment 1869
In triangle ABC, with sides a, b, and c opposite the respective angles A, B, and C, the law of tangents states:

(a-b)/(a+b)=tan((A-B)/2)/tan((A+B)/2)….(1)

Similarly for other sides,

(b-c)/(b+c)=tan((B-C)/2)/tan((B+C)/2)….(2)

(c-a)/(c+a)=tan((C-A)/2)/tan((C+A)/2) …..(3)


if a = 321, b = 234 and angle C = 71°

(234-c)/(234+c)=tan((B-71)/2)/tan((B+71)/2)
c = 234 sin(71) (1/sin(B))...........(2a)

(c-321)/(c+321)=tan((71-A)/2)/tan((71+A)/2)
c = 321 sin(71) (1/sin(A))................(3a)

from (2a) and (3a) we have

234 sin(71) (1/sin(B))=321 sin(71) (1/sin(A))..........simplify

234/sin(B)=321/sin(A)

234sin(A)=321sin(B)

sin(A)=(321/234)sin(B)
sin(A)=(107/78)sin(B)
A=sin^-1((107/78)sin(B))

go to
c = 321 sin(71) (1/sin(A))................(3a), substitute A
c = 321 sin(71) (1/sin(sin^-1((107/78)sin(B))))
c = 321sin(71)(107/78) sin(B)..........(3b)

from (2a) and (3b) we have

234 sin(71) (1/sin(B))=321sin(71)(107/78) sin(B).......solve for B

234/sin(B)=321(107/78) sin(B)
sin^2(B)=234/(321(107/78))
sin^2(B)=6084/11449

sin(B)=sqrt(6084/11449)

sin(B)=0.7289719626168224

B=sin^-1(0.7289719626168224)
B=46.80027915261779°

since given C = 71°, then angle A will be

A=180-( 71°+46.80027915261779°)
A=180°-( 117.80027915261779° )
A=62.19972084738221°

go to
c = 321 sin(71) (1/sin(A)), substitute A
c = 321 sin(71) (1/sin(62.19972084738221))
c = 343.11414209938454

angle A, angle B and little c to the nearest tenth.
A=62.2°
B=46.8°
c = 343.1

Wow! This is complicated. I can find angles A and B and side c without the law of tangents. So, I will stick to the law of sines and law of cosines. Why complicate my life? Right?