Right Triangle

**soni** · June 29th 2010, 11:13 PM

In a right angled triangle ABC shown below, tan(angle ABC) = 1/2 and AB(hypotense)=square root of 20cm. Using simultaneous equations, find the length of AC(opposite) and BC(adjacent).

**Prove It** · June 29th 2010, 11:23 PM

Originally Posted by soni

In a right angled triangle ABC shown below, tan(angle ABC) = 1/2 and AB(hypotense)=square root of 20cm. Using simultaneous equations, find the length of AC(opposite) and BC(adjacent).

You are told that $\tan{\theta} = \frac{1}{2}$ .

So $\frac{O}{A} = \frac{1}{2}$ .

You also have, by Pythagoras' Theorem

$O^2 + A^2 = H^2$

$O^2 + A^2 = (\sqrt{20})^2$

$O^2 + A^2 = 20$ .

So solve $\frac{O}{A} = \frac{1}{2}$ and $O^2 + A^2 = 20$ simultaneously.

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