If and are acute angles such that , then show that .

Understand the Question Given that and are acute angles satisfying . In any right-angled triangle, . To prove in general, we consider two separate right-angled triangles containing and respectively, and prove they are similar using the SSS (Side-Side-Side) similarity criterion. Step 1 · Relate Sides using Cosine Consider two right-angled triangles right-angled at , and right-angled at . From the definition of cosine Given Rearranging the terms This gives Step 2 · Prove Triangle Similarity By Pythagoras theorem Taking the ratio of the third sides Therefore By the SSS similarity criterion, . Since corresponding angles of similar triangles are equal Answer > Common Mistakes Assuming a Single Triangle Only: While proving the result in a single right triangle where is valid, the general proof requires considering two independent triangles using similarity. Skipping Third Side Proportionality: Directly claiming triangles are similar without evaluating via the Pythagoras theorem.

Introduction to Trigonometry | Exercise 8.1

Question 6

If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$ , then show that $\angle A = \angle B$ .

Solution

Understand the Question

Given that $\angle A$ and $\angle B$ are acute angles satisfying $\cos A = \cos B$ .
In any right-angled triangle, $\cos\theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$ .
To prove $\angle A = \angle B$ in general, we consider two separate right-angled triangles containing $\angle A$ and $\angle B$ respectively, and prove they are similar using the SSS (Side-Side-Side) similarity criterion.

Step 1 · Relate Sides using Cosine

Consider two right-angled triangles $\triangle APQ$ right-angled at $P$ , and $\triangle BCD$ right-angled at $C$ . Diagram 1

From the definition of cosine $\cos A = \dfrac{\text{Adjacent side}}{\text{Hypotenuse}} = \dfrac{AP}{AQ}$

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

Given $\cos A = \cos B$ $\dfrac{AP}{AQ} = \dfrac{BC}{BD}$

Rearranging the terms $\dfrac{AP}{BC} = \dfrac{AQ}{BD} = k$

This gives $AP = k \cdot BC$ $AQ = k \cdot BD$

Step 2 · Prove Triangle Similarity

By Pythagoras theorem $PQ = \sqrt{AQ^2 - AP^2}$ $CD = \sqrt{BD^2 - BC^2}$

Taking the ratio of the third sides

\begin{aligned} \dfrac{PQ}{CD} &= \dfrac{\sqrt{AQ^2 - AP^2}}{\sqrt{BD^2 - BC^2}} \\[0.6em] &= \dfrac{\sqrt{(k \cdot BD)^2 - (k \cdot BC)^2}}{\sqrt{BD^2 - BC^2}} \\[0.6em] &= \dfrac{\sqrt{k^2 \cdot BD^2 - k^2 \cdot BC^2}}{\sqrt{BD^2 - BC^2}} \\[0.6em] &= \dfrac{\sqrt{k^2 (BD^2 - BC^2)}}{\sqrt{BD^2 - BC^2}} \\[0.6em] &= \dfrac{k \sqrt{BD^2 - BC^2}}{\sqrt{BD^2 - BC^2}} \\[0.6em] &= k \end{aligned}

Therefore $\dfrac{AP}{BC} = \dfrac{AQ}{BD} = \dfrac{PQ}{CD} = k$

By the SSS similarity criterion, $\triangle APQ \sim \triangle BCD$ .

Since corresponding angles of similar triangles are equal $\angle A = \angle B$

Answer

$\angle A = \angle B$

Common Mistakes

Assuming a Single Triangle Only: While proving the result in a single right triangle where $\angle A + \angle B = 90^\circ$ is valid, the general proof requires considering two independent triangles using similarity.
Skipping Third Side Proportionality: Directly claiming triangles are similar without evaluating $\dfrac{PQ}{CD} = k$ via the Pythagoras theorem.

Question 6

More questions in Exercise 8.1