# Trigonometry

#### Elize

In a triangle ABC A=90 degree
And (b-c)/(b+c)= 1/sqrt(3)
Find angle B??

#### skeeter

Math Team
$\dfrac{b-c}{b+c} = \dfrac{1}{\sqrt{3}} \implies \dfrac{b}{c} = 2+\sqrt{3}$

$B = \arctan\left(\dfrac{b}{c}\right)$

1 person

#### skipjack

Forum Staff
$$\displaystyle \tan\left(30^\circ\right) = \frac{1}{\sqrt3} = \frac{b/c - 1}{1 + b/c} = \tan\left(B - 45^\circ\right) \implies B = 75^\circ$$

greg1313

#### greg1313

Forum Staff
$$\displaystyle \tan\left(30^\circ\right) = \frac{1}{\sqrt3} = \frac{b/c - 1}{1 + b/c} = \tan\left(B - 45^\circ\right) \implies B = 75^\circ$$
Nicely done. As inspired,

$$\displaystyle b-c=1$$

$$\displaystyle b+c=\sqrt3$$

$$\displaystyle 2b=\sqrt3+1$$

$$\displaystyle b=\frac{\sqrt3+1}{2}$$

Similarly,

$$\displaystyle c=\frac{\sqrt3-1}{2}$$

Using the Pythagorean theorem,

$$\displaystyle a=\sqrt2$$

Then,

$$\displaystyle \sqrt2\sin(B)=\frac{\sqrt3+1}{2}$$

$$\displaystyle \sin(B)=\frac{\sqrt3}{2}\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\frac12=\sin(60^\circ+45^\circ)=\sin(105^\circ)=\sin(75^\circ)$$

As triangle $ABC$ is right, $B=75^\circ$.