In the angle shown below, for different points on the slanted line, as the distance from the vertex of the angle changes, the height from the horizontal line also changes.

i) Prove that height is proportional to the distance.

ii) Calculate the constant of proportionality for 30°, 35° and 60° angles.

i). The given angle in question figure is reproduced in below figure.

Let the blue object which moves along the slanting line is marked as Q and perpendicular is dropped from Q to the horizontal leg of the given angle. The foot of perpendicular is marked as P.

So, we get a triangle Δ APQ

There is another right-angle ΔABC. This is our base triangle. That is, we are going to calculations based on Δ ABC:

We assume that BC is fixed at its position and also that all sides of Δ ABC are known.

But PQ is not fixed. Because, Q can be at any point along the slanting line.

Now, we find relation between Δ ABC and Δ APQ:

∠ A is denoted as x° (It is same for both the Δ)

∠ P and ∠ B are 90°

∠ c and ∠ Q will both be equal to (90 – x) °

So, both the triangles have the same angles. Therefore, they are similar.

i.e. Δ ABC ≅ Δ APQ

⇒

From the above,

∵ BC and PQ are heights and AC and AQ are distance.

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⇒

⇒

is constant bcause Δ ABC is fixed.

⇒ AQ = k × PQ

That means, the distance of Q from the vertex A is proportional to the height from Q from the horizontal line.

ii). In this part we explore the cases when the angle x at vertex A is 30°, 60° and 45°.

First, we will take 30°.

A base triangle Δ ABC is drawn such that AC = 2cm, BC = 1cm and AB = √3

As we see in figure, here also Δ ABC and Δ APQ are similar

So, we get:

∵ BC and PQ are heights and AC and AQ are distance.

⇒

⇒

⇒

⇒ AQ = 2 PQ

In the above result, ‘2’ is constant. So AQ is proportional to PQ.

That means the distance of Q from the vertex A is proportional to the height Q from the horizontal line.

Now take x°= 60°,

A base triangle Δ ABC is drawn such that AC = 2cm, AB = 1cm and BC = √3

As we see in figure, here also Δ ABC and Δ APQ are similar

So, we get:

∵ BC and PQ are heights and AC and AQ are distance.

⇒

⇒

⇒

In the above result, is constant. So AQ is proportional to PQ.

That means the distance of Q from the vertex A is proportional to the height Q from the horizontal line.

Now take x°= 35°,

A base triangle Δ ABC is drawn such that AC = √2cm, AB = 1cm and BC = 1cm

As we see in figure, here also Δ ABC and Δ APQ are similar

So, we get:

∵ BC and PQ are heights and AC and AQ are distance.

⇒

⇒

⇒

⇒ AQ = √2 PQ

In the above result, is constant. So AQ is proportional to PQ.

That means the distance of Q from the vertex A is proportional to the height Q from the horizontal line.