# Show that b=3c

• January 27th 2011, 11:16 AM
razemsoft21
Show that b=3c
$Triangle\: ABC\: with\: A=60$

$if\: \sin A=\sqrt 7 \sin C$

show that $b=3c.$
• January 27th 2011, 11:50 AM
Ithaka
Quote:

Originally Posted by razemsoft21
$Triangle\: ABC\: with\: A=60$

$if\: \sin A=\sqrt 7 \sin C$

show that $b=3c.$

A=60, sin A= $\frac{\sqrt3}{2}$

sin C= $\frac{sin A}{\sqrt 7}$

sin C= $\frac{\sqrt 3}{2 * \sqrt 7}$

sin C= $\frac{\sqrt 21}{14}$

from here, work out cos C

cos C = $\frac{5*\sqrt 7}{14}$ (I am taking cos C to be positive,
otherwise angle B would come out reflex)

since A+B+C = 180, B=180-(A+C), sin B=sin (A+C)=sin A * cos C + cos A * sin C

= $\frac{3*\sqrt 21} {14}$

Now use sine rule in the triangle ABC, you will get b=3c
• January 27th 2011, 12:31 PM
razemsoft21
Nice job, thanks a lot
b=3c