# Thread: In any given triangle prove that...

1. ## In any given triangle prove that...

In any given triangle prove that...

$tan\frac{1}{2}(B-C) = tan(45\,^{\circ}-\theta)cot\frac{1}{2}A$
where $tan\theta=\frac{c}{b}$

find, $\frac{1}{2}(B-C)$ if $b=321, c=436, A=119\,^{\circ}15'$

the answer is... $-5\,^{\circ}5'; (\theta = 53\,^{\circ}38')$

not sure how they got this...

2. Hello bigwave
Originally Posted by bigwave
In any given triangle prove that...

$tan\frac{1}{2}(B-C) = tan(45\,^{\circ}-\theta)cot\frac{1}{2}A$
where $tan\theta=\frac{c}{b}$

find, $\frac{1}{2}(B-C)$ if $b=321, c=436, A=119\,^{\circ}15'$

the answer is... $-5\,^{\circ}5'; (\theta = 53\,^{\circ}38')$

not sure how they got this...
A few preliminary results:
$\cot\tfrac12A=\cot\tfrac12(180^o-[B+C])=\cot(90^o-\tfrac12[B+C])=\tan\tfrac12(B+C)$ (1)

$\tan 45^o=1$ (2)

Using the Sine Rule: $\frac{c}{b}=\frac{\sin C}{\sin B}=\frac{\sin \tfrac12C\cos \tfrac12C}{\sin \tfrac12B\cos\tfrac12 B}$ (3)
Then, starting with the LHS:
$\tan(45^o-\theta)\cot\tfrac12A$
$=\frac{\tan45^o-\tan\theta}{1+\tan45^o\tan\theta}\cdot\tan\tfrac12 (B+C)$ using (1)

$=\frac{1-\dfrac{c}{b}}{1+\dfrac{c}{b}}\cdot\tan\tfrac12(B+C )$ using (2)

$=\frac{1-\dfrac{\sin \tfrac12C\cos \tfrac12C}{\sin \tfrac12B\cos\tfrac12 B}}{1+\dfrac{\sin \tfrac12C\cos \tfrac12C}{\sin \tfrac12B\cos\tfrac12 B}}\cdot\frac{\tan\tfrac12B+\tan\tfrac12C}{1-\tan\tfrac12B\tan\tfrac12C}$ using (3)

$=\frac{\sin \tfrac12B\cos\tfrac12 B-\sin \tfrac12C\cos \tfrac12C}{\sin \tfrac12B\cos\tfrac12 B+\sin \tfrac12C\cos \tfrac12C}\cdot\frac{\tan\tfrac12B+\tan\tfrac12C}{ 1-\tan\tfrac12B\tan\tfrac12C}$

$=\frac{\tan \tfrac12B-\tan \tfrac12C}{\tan \tfrac12B+\tan \tfrac12C}\cdot\frac{\tan\tfrac12B+\tan\tfrac12C}{ 1-\tan\tfrac12B\tan\tfrac12C}$

$=\tan\tfrac12(B-C)$
I agree with the numerical answer given. This is a straightforward use of a calculator, using the result we've just proved. Is there a problem with this?