# How to find triangle height?

I need to know the height ($h$) of a triangle with two unknown angles ($\alpha$ and $\beta$) and the known length of two sides $AB$ and $BC$.

Is it possible to have that value of $h$ (height)?

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Which sides and which angles do you know - the diagram is confusing and suggests, contrary to text that $\alpha$ and $\beta$ are known. –  Mark Bennet Jul 3 '11 at 20:04
All angles are unknown and two lengths of sides are known. I realize that we can't find h with theses conditions. –  elbaid Jul 3 '11 at 20:11
all you can do then is bound the height. It can't be less than zero. And, as you've drawn it, it can't be greater than BC. –  Mark Bennet Jul 3 '11 at 20:21
@elbait: A triangle is defined if and only if: 1) two angles and a side are known; 2) two sides and the angle between them are kown; 3) three sides are known. –  Américo Tavares Jul 3 '11 at 20:27
@Américo Tavares: and if two sides and an angle not between them are known the cosine formula (and trying to draw a scale model) give at most two possibilities. –  Mark Bennet Jul 3 '11 at 20:41

Your post says that the angles $\alpha$ and $\beta$ are unknown. I assume this means that only $AB$ and $BC$ are known. Remove the line $AC$ from your diagram. Then you can turn the line $BC$ around the "hinge" at $B$, changing the height. So the answer is that the height cannot be determined.

The only thing you can say is that the height is greater than $0$ and less than or equal to $BC$.

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"Unknown" was added after a couple of us answered. See revision history... –  amWhy Jul 3 '11 at 20:00
@amWhy: I did wonder briefly why the one answer I had seen assumed that $\beta$ is known. Didn't think of looking at the revision history. –  André Nicolas Jul 3 '11 at 20:08

If you let D denote the point of intersection of the perpendicular point C to segment AB (so that the length of CD = h), then we have a right triangle BCD ($\angle BDC = 90^\circ = \pi/2$ radians),

then you have that $\sin(\beta)$ = opposite side / hypotenuse = $\displaystyle \frac{h}{|BC|}$

Then solve for h: $h = \sin(\beta)\times |BC|$.

Since you have the values for $\beta$ and for $|BC|$ (the length of segment BC), simply compute h at those values.

Note: this answer was posted when the question stated $\alpha, \beta$ were known. With only two sides of the triangle known (and that the line perpendicular to AB, and intersecting C, forms divides ABC into two right triangles), we still don't have enough information to determine height (h).

See Americo's comment above for a list of values (side lengths and/or angles) necessary to define a triangle. (And his subsequent comment re: "case 4")!

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since you rolled back my edition, could you edit $|BC| by yourself? – Ilya Jul 3 '11 at 21:01 @Gortaur: I'm not sure what you mean "rolled back"...did you try editing$|BC|? Perhaps it was while I was editing (adding Note, and then again to refer to Americo's comment above. Anyway, sorry if that "cut you (or your edit) off"...I didn't manually roll back or anything...Also Thanks! for pointing out the one-sided $! =) – amWhy Jul 3 '11 at 21:10 @Gortaur: I just looked at the revision history, and see that it does look as though I must have inadvertently clicked "rollback"? somehow? at some point. I wouldn't object in the least to edits fixing my typos, etc....! – amWhy Jul 3 '11 at 21:18 From the definition of$\sin$function,$h = |BC|\sin\beta$. Even if you know only$|AB| = a$(which seems to be the case from your picture) then:$h = ka\cos\alpha$and$h = (1-k)a\cos\beta$for some$k<1$. This$k$is a proportion in which$h$divides$AB\$. So, we have to equation on two variables: Jejeje $$k = \frac{\cos\beta}{\cos\alpha+\cos\beta}$$ and $$h = \frac{\cos\alpha\cos\beta}{\cos\alpha+\cos\beta}a.$$

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Sorry, my question was badly put... this is unknown angles alpha and beta. –  elbaid Jul 3 '11 at 19:48