Find the formula for the height of a triangle using ONLY the base length and two angles.

Question

This is the question I have been presented with.

I have tried looking at law of sines as a solution, but that requires the use of the third angle. (Of course, It can be done but I don't think the problem wants me to use it?)

Is it possible to create a formula to find the height with only the given variables? Am I interpreting the problem wrongly or something?

Answer 1

Decompose your base into

$$\ell = \ell_1+\ell_2 \tag{1}$$

As we can write :

$$\dfrac{d}{l_1}=\tan \alpha \ \ \text{and} \ \ \dfrac{d}{l_2}=\tan \beta$$

we get

$$\ell_1+\ell_2=\dfrac{d}{\tan \alpha}+\dfrac{d}{\tan \beta}$$

Using (1), one obtains :

$$d=\dfrac{\ell}{1/\tan(\alpha)+1/\tan(\beta)}$$

Answer 2

Use SOH CAH TOA on the two right-angle triangles you have to create an equation.

Let's call the point P1 '$A$', the point P2 '$B$' and the third point of the triangle '$C$'. Call the point where the perpendicular from $C$ hits the line $AB$ '$D$'.

Then the length of $AD$ can be expressed in terms of $d$ and $\tan\alpha$. Also, the length of $BD$ can be expressed in terms of $d$ and $\tan\beta$. Then you know that $AD+BD=l$. You then rearrange this equation to make $d$ the subject of the formula.

Answer 3

You may just set up the formula as

$$l= d\cot\alpha + d\cot\beta$$

which means that the base length is made of two segments, which are the sides of two right triangles and measured as $d\cot\alpha$ and $d\cot\beta$, respectively.

As a result, the distance is simply,

$$d=\frac{l}{\cot\alpha + \cot\beta}$$

Answer 4

Call $\;x\;$ the distance from $\;P_1\;$ to the intersection of the red line with the shore, so that the distance from this point to $\;P_2\;$ is $\;\ell-x\;$ .

Thus, on the left right triangle we get $\;\tan\alpha=\cfrac dx\;$ , and on the right right atriangle $\;\tan\beta=\cfrac d{\ell-x}\;$, and then

$$x\tan\alpha=d=(\ell-x)\tan\beta\implies\left(\tan\alpha+\tan\beta\right)x=\ell\tan\beta\implies x=\frac{\ell\tan\beta}{\tan\alpha+\tan b\beta}$$

and now go back to the first relation on the left right triangle to get your expression...

Answer 5

Hint:

Denote $a$ thside of the triangle opposite of angle $\alpha$, $b$ the side opposite to angle $\beta$ and apply the law of sines: $$\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{\ell}{\sin(\alpha+\beta)}$$ and give the expressions of $\sin\alpha$ and $\sin\beta$ from the right triangles determined by the altitude.

You should obtain the very simple formula: $$d=\frac{\ell\sin\alpha\sin\beta}{\sin(\alpha+\beta)}.$$

Answer 6

Let $\gamma$ be the third angle and let $b$ be the length of the side opposite $\beta$. Then $$\gamma=180^\circ-\alpha-\beta$$ $$\sin\gamma=\sin(180^\circ-(\alpha+\beta))=\sin(\alpha+\beta)$$ $$\frac b{\sin\beta}=\frac l{\sin\gamma}$$ $$d=l\sin\alpha$$

Putting those together we get $$d=\frac{l\sin\alpha\sin\beta}{\sin(\alpha+\beta)}$$