I was playing a flight simulator game and attaching some cargo on the ground to a helicopter, which was hovering above ground level. There were 3 individual pieces of cargo at different distances from the helicopter. After thinking about it for a while, I came up with this nice little puzzle:
A Helicopter hovering at some height is collecting 3 cargo bags on the ground. Bag A is 1 meter away from the helicopter (horizontal distance). Bag B is 2 meters away (horizontal distance) and Bag C is 3 meters away (horizontal distance). The harness wire connecting the helicopter and the cargo bags form some angle of elevation. If the sum of the angles of elevation of the cargo bag is $90^\circ$, what is the height of the helicopter?
I thought about this problem for a while and I came up with one solution to this, as I'll share my attempt below. I want to ask, are there any other ways to approach this with? Perhaps with less trigonometry? Please share your attempts too!
Here's how I approached it:
Let the height of the helicopter be $x$ meters, and I'll refer to the angles of elevation of the wired connecting Bags A, B and C with these letters respectively. now let's take the tangent of every angle, we'll have:
$$\tan A=\frac{x}{1}$$ $$\tan B=\frac{x}{2}$$ $$\tan C=\frac{x}{3}$$
We know that:
$$A+B+C=\frac{\pi}{2}$$ $$A+B=\frac{\pi}{2}-C$$ $$\tan (A+B)=\cot C$$ $$\frac{\tan A +\tan B}{1-\tan(A)\tan(B)}=\frac{1}{\tan C}$$ $$\tan(A)\tan(C) +\tan(B)\tan(C)+\tan(A)\tan(B)=1$$
Now substituting the values for tangent that we derived above, we get:
$$\frac{x^2}{3}+\frac{x^2}{6}+\frac{x^2}{2}=1$$ $$2x^2+x^2+3x^2=6$$ $$6x^2=6$$
Therefore the height of the helicopter is $x=1$ meters. Are there any other ways to solve this?
Alternative approach:
(Actually not that different from the original poster's, and perhaps not as elegant as the original poster's).
Let $a,b,c$ denote the three angles.
Then, $~\tan(a) = (h), ~\tan(b) = (h/2), ~\tan(c) = (h/3).$
You need to determine $h$ so that $\tan(a + b + c) = \infty.$
$$\tan(a+b) = \frac{\frac{3h}{2}}{1 - \frac{h^2}{2}} = \frac{3h}{2 - h^2}.$$
$$\tan(a+b+c) = \frac{\frac{3h}{2 - h^2} + \frac{h}{3}}{1 - \left[ ~\frac{3h}{2 - h^2} \times \frac{h}{3} ~\right]}$$
$$= \frac{\frac{\text{something}}{(2-h^2) \times 3}}{\frac{6 - 6h^2}{(2-h^2) \times 3}} = \frac{\text{something}}{6 - 6h^2}.$$
So, $h$ must be a positive real number that satisfies $6 - 6h^2 = 0.$ This implies that $h$ must equal $1$.