### 11.4.1 Measuring Triangles

This subsection is not completed yet, but in the meantime here are some practice questions.

Remember:
The special angles that you need to know, with their sines, cosines, and tangents are:

 $\theta$ $\sin\theta$ $\cos\theta$ $\tan\theta$ $30^\circ=\pi/6$ $\dfrac{\sqrt1}{2}$ $\dfrac{\sqrt3}{2}$ $\dfrac{\sqrt1}{\sqrt3}=\dfrac{1}{\sqrt3}$ $45^\circ=\pi/4$ $\dfrac{\sqrt2}{2}$ $\dfrac{\sqrt2}{2}$ $\dfrac{\sqrt2}{\sqrt2} =1$ $60^\circ=\pi/3$ $\dfrac{\sqrt3}{2}$ $\dfrac{\sqrt1}{2}$ $\dfrac{\sqrt3}{\sqrt1}=\sqrt3$

When written in this way, the sines make a lovely pattern of $\sqrt1/2$, $\sqrt2/2$, $\sqrt3/2$, with the cosines in the opposite order. You can work out the tangents by dividing sine by cosine. (I’m using $\sqrt1/2$ to make the pattern clear, but we would of course use $1/2$ instead.)

Alternatively, you can work it out by sketching the special triangles: Or if you want to use a calculator, you can learn to recognise $0.866\cdots$ as $\sqrt3/2$ and $0.707\cdots$ as $\sqrt2/2$, or you can get the exact value by squaring the answer, for example $\sin60^\circ$ will give $0.866\cdots$, and if you square this, you’ll get $3/4$, whose square root is $\sqrt3/2$.