# trig

• Jul 28th 2007, 07:38 AM
harry
trig
In a right triangle given a=32.8ft and A= 28 deg 30’15” , solve for b and c
• Jul 28th 2007, 09:55 AM
DivideBy0
Quote:

Originally Posted by harry
In a right triangle given a=32.8ft and A= 28 deg 30’15” , solve for b and c

$\left(28+\frac{30}{60}+\frac{15}{360} \right)^\circ =\frac{685 }{24}^\circ$

Using the Sine Rule:

$\frac{a}{\sin A}=\frac{b}{\sin B} = \frac{c}{\sin C}$

$\frac{32.8}{\sin(\frac{685}{24})}=\frac{c}{\sin90}$

$c \approx 68.65$

Use pythag to get b:

$b=\sqrt{68.65^2+32.8^2}\approx 76.08$

Then use the sine rule again to solve for angle B.

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