Calculate height of triangle given angle and base
Mia Lopez 
I need to calculate the height of $x$ in my triangle and I know the base as well as the inner angles. I've used the following equation to try calculate it based off this answer here:
How do you find the height of a triangle given $3$ angles and the base side? Image given.
The equation that I got based off that answer was this:
$$x = \frac{1000 \sin(90) \sin(5)}{ \sin(85)}$$
this gives me a value of around $4000$ which can't be right so not sure where i've gone wrong here?
$\endgroup$ 22 Answers
$\begingroup$Recall that $\tan 5 ^{\circ}=\frac{x}{1000}$
$\endgroup$ $\begingroup$It looks like you were trying to use the Law of Sines. That rule is often presented as: $$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c},$$ which makes it look like one formula involving $a$ and $b$ and $c$, but it is only applied to two at a time. In this case, we're not worried about $a$, so we can ignore both it and the right angle. Then we have: $$\frac{\sin 5^\circ}{x}=\frac{\sin 85^\circ}{1000},$$ which can be solved for $x$.
Keep in mind, though, the Law of Sines is not the easiest way to approach this problem. When the triangle has a right angle, we can directly relate sides and angles using the right-triangle definitions of sine, cosine and tangent:
- $\sin \theta=\frac{\text{opposite}}{\text{hypotenuse}}$
- $\cos \theta=\frac{\text{adjacent}}{\text{hypotenuse}}$
- $\tan \theta=\frac{\text{opposite}}{\text{adjacent}}$
You pick the one that applies to your problem, and plug in: Here we have one leg of the right triangle, and we want the other, so we use tangent: $$\tan 5^\circ=\frac{x}{1000}$$
If you're working with a triangle that's not a right triangle, then $\sin$ is often the most useful function for finding heights, but that's another question.
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