Question 6

Acute Angle: An angle strictly between 0° and 90°. In a right-angled triangle, both non-right angles are acute.

Right-Angled Triangle: A triangle with one 90° angle. For any acute angle in it:

$\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$

Key idea: If $\cos A = \cos B$, the ratios of adjacent to hypotenuse are equal in both triangles. This means the triangles are proportional — and by SSS similarity, corresponding angles must be equal, giving $\angle A = \angle B$.

SSS Similarity: If all three pairs of corresponding sides of two triangles are proportional, the triangles are similar and their corresponding angles are equal.

We will use two right-angled triangles to show the equality of angles.

Step 1 — Constructing Right Triangles

Let's consider two right-angled triangles. Let $\triangle APQ$ be right-angled at $P$. Let $\angle A$ be one of its acute angles. Let $\triangle BCD$ be right-angled at $C$. Let $\angle B$ be one of its acute angles.

Step 2 — Relating Sides using Cosine

We know the definition of cosine in a right triangle. $\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AP}{AQ}$

$\cos B = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{BC}{BD}$

We are given that $\cos A = \cos B$. So, we can write: $\frac{AP}{AQ} = \frac{BC}{BD}$

Let's rearrange this proportion. $\frac{AP}{BC} = \frac{AQ}{BD}$

Let this common ratio be $\mathbf{k}$. $\frac{AP}{BC} = \frac{AQ}{BD} = k$

From this, we get: $AP = k \cdot BC$ $AQ = k \cdot BD$

Step 3 — Proving Triangle Similarity

Now, let's find the third side of each triangle using the Pythagorean theorem. In $\triangle APQ$: $PQ = \sqrt{AQ^2 - AP^2}$

In $\triangle BCD$: $CD = \sqrt{BD^2 - BC^2}$

Let's find the ratio of these third sides. $\frac{PQ}{CD} = \frac{\sqrt{AQ^2 - AP^2}}{\sqrt{BD^2 - BC^2}}$

Substitute the expressions for $AQ$ and $AP$ in terms of $k$. $\frac{PQ}{CD} = \frac{\sqrt{(k \cdot BD)^2 - (k \cdot BC)^2}}{\sqrt{BD^2 - BC^2}}$

$= \frac{\sqrt{k^2 \cdot BD^2 - k^2 \cdot BC^2}}{\sqrt{BD^2 - BC^2}}$

$= \frac{\sqrt{k^2 (BD^2 - BC^2)}}{\sqrt{BD^2 - BC^2}}$

$= \frac{k \sqrt{BD^2 - BC^2}}{\sqrt{BD^2 - BC^2}}$

$= k$

So, we have established that: $\frac{AP}{BC} = \frac{AQ}{BD} = \frac{PQ}{CD} = k$

This means that the corresponding sides of $\triangle APQ$ and $\triangle BCD$ are proportional. Therefore, $\triangle APQ \sim \triangle BCD$ by the SSS (Side-Side-Side) similarity criterion.

Step 4 — Concluding Equality of Angles

Since the two triangles $\triangle APQ$ and $\triangle BCD$ are similar, their corresponding angles must be equal. The angle corresponding to $\angle A$ in $\triangle APQ$ is $\angle B$ in $\triangle BCD$. Hence, we can conclude that:

$\boxed{\angle A = \angle B}$