Trigonometry

Feb 2018
63
3
Iran
In a triangle ABC A=90 degree
And (b-c)/(b+c)= 1/sqrt(3)
Find angle B??
 

skeeter

Math Team
Jul 2011
3,364
1,855
Texas
$\dfrac{b-c}{b+c} = \dfrac{1}{\sqrt{3}} \implies \dfrac{b}{c} = 2+\sqrt{3}$

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

Forum Staff
Dec 2006
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\(\displaystyle \tan\left(30^\circ\right) = \frac{1}{\sqrt3} = \frac{b/c - 1}{1 + b/c} = \tan\left(B - 45^\circ\right) \implies B = 75^\circ\)
 
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greg1313

Forum Staff
Oct 2008
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London, Ontario, Canada - The Forest City
\(\displaystyle \tan\left(30^\circ\right) = \frac{1}{\sqrt3} = \frac{b/c - 1}{1 + b/c} = \tan\left(B - 45^\circ\right) \implies B = 75^\circ\)
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, $B=75^\circ$.