Right Triangle Trig: Why are the Angles the Same?

Question

On Khan Academy, the problem involves this picture of a triangle:

In the problem, we must solve for the theta angle, so in the explanations, the triangle was broken up into a square so that there is the same angle theta on another triangle.

Here, how do you explain the same angles in this triangle and the one in triangle KGJ? Perhaps an idea being a certain rule of right triangle trig to be followed.

Answer 1

It must be noted that they are corresponding angles. Under the "Corresponding Angles Axiom", we state that

"When a transversal (here, the line $t$) cuts two parallel lines, then the corresponding angles ($\alpha$ and $\alpha_1$) are equal."

This is an axiom, and can be visually shown by superposition.

In your case, the line through $K$ at $HF$ is parallel to the line $GF$, and line $GH$ is the transversal. Hence, the angles under consideration are equal.

Answer 2

Call the vertex with $90^{\circ}$ between side lengths $(5,7)$ as $L$. The angle $ HGL$ should not be labelled as $\theta$ already.

The angles $KLF,LFG$ add up to $180^{0},\;. $ So lines $FG,LK$ are parallel.

$LK$ should make the same angle to the transversal cutting line $HKG$ by virtue of parallelism property of corresponding angles.

Answer 3

Euclid's geometry states that equal angles are formed by one line such as HG crossing parallel lines such as FG and the horizontal that includes K above it. This means that the angle $\theta$ near $K$ is the same as the angle $\theta$ near G. Since all angles of all three triangle are equal, the triangles are similar.

As for solving, we know that $$HK=\sqrt{5^2+7^2}\approx 8.6$$ $$FG=\frac{7}{5}\times 12=\frac{7+60}{5}\approx 13.4$$ $$HG=\sqrt{12^2+13.4^2}\approx 17.99$$ $$\theta = \tan^{-1}(5/7)\approx 0.62 r\approx 35.54^\circ $$