I had this question in my test:- Authority wants to construct a slide in a city park for children. Authority prefers the top of the slide at a height of 4m above the ground and inclined at an angle of 30° with the ground.
So one of the subparts of this question was:- If AB + BC = 25m and AC = 5m, then the value of BC is:- (A) 25m (B) 15m (C) 10m (D) 12m
Actual solution:
So the solution was by assuming BC = x and AB = 25 - x. Then by applying Pythagoras Theroem we have: $(25-x)^2 = x^2 + 5^2$ and then after solving for x we get x = BC = 12m
My Approach: I applied sine for angle B : $\sin(B) = \sin(30°) = \dfrac{1}{2} = \dfrac{AC}{AB} = \dfrac{5}{AB} \\ \implies AB = 10m$
Then by using the relation: $AB + BC = 25m$, I got $BC = 15m.$
My Problems:
Why do the two answers differ?
In the way I solved, the sides are: 5, 10 and 15, which is not a Pythagoream Triplet. How is this possible since the slide should be perpendicular to the ground, and even more, trigonometric ratios exist for right triangles only.
In my answer why is the side BC which is a leg has more length than the hypotenuse AB
Whoever wrote the "test" was not thinking clearly enough about trigonometry and the Pythagorean theorem. The $\textbf{Actual solution}$ suggests $\quad (25-x)^2 = x^2 + 5^2 \implies 600-50x=0 \implies x=12\quad$ but, with $\angle ABC=30^\circ,\space$ none of the sides are integer multipls of $3$ or $5$.
The first paragraph and the picture describe half of an equilateral triangle and $\textit{that}$ half has dimensions of $(4\sqrt{3},4,8)\approx (6.93,\space 4,\space 8) $ where $4\sqrt{3}$ is the "altitude" <sarcasm> or the $x$-axis running along the ground.
This means it is not a Pythagorean triple where all sides are integers. The Pythagorean theorem does apply in
$$BC^2+AC^2=AB^2\\ (4\sqrt{3})^2+4^2=8^2$$