The diagram shows triangle ABC in which AC=4cm, BC=5cm and ACB=60 degrees.

a) Show that AB = Square root of 21

b) find the value of sin (angle ABC) in the form k(square root)7 where k is an exact fraction

Anyone know how to do this?

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- Feb 3rd 2008, 08:19 AMColin_mSine and Cosine rules question
The diagram shows triangle ABC in which AC=4cm, BC=5cm and ACB=60 degrees.

a) Show that AB = Square root of 21

b) find the value of sin (angle ABC) in the form k(square root)7 where k is an exact fraction

Anyone know how to do this? - Feb 3rd 2008, 08:31 AMbobak
For the first make you use the cosine rule

$\displaystyle AB^2 = 4^2 + 5^2 - 2 \cdot 4 \cdot 5 \cdot \cos 60$

can you finish that off ?

for the second part use the sine rule

$\displaystyle \frac{4}{ \sin ABC} = \frac{ \sqrt {21}}{ \sin 60}$

note $\displaystyle \sin 60 = \frac{ \sqrt {3} }{2}$ - Feb 3rd 2008, 08:47 AMColin_m
yep, thanks bobak.