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?
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?
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 $\displaystyle \sin 60 = \frac{ \sqrt {3} }{2}$