Question 5

Right-Angled Triangle: For any acute angle $\theta$, the sides are:

• Opposite — side facing $\theta$
• Adjacent — side next to $\theta$
• Hypotenuse — longest side, opposite the 90° angle

All six trigonometric ratios:

$\sin\theta = \frac{\text{Opp}}{\text{Hyp}}, \quad \cos\theta = \frac{\text{Adj}}{\text{Hyp}}, \quad \tan\theta = \frac{\text{Opp}}{\text{Adj}}$

$\csc\theta = \frac{\text{Hyp}}{\text{Opp}}, \quad \sec\theta = \frac{\text{Hyp}}{\text{Adj}}, \quad \cot\theta = \frac{\text{Adj}}{\text{Opp}}$

Pythagoras Theorem: $\text{Hypotenuse}^2 = \text{Opposite}^2 + \text{Adjacent}^2$

Since $\sec\theta = \frac{13}{12}$, we take hypotenuse $= 13k$ and adjacent $= 12k$, then find opposite using Pythagoras.

We can use a right-angled triangle to find the lengths of its sides.

Step 1 — Find the unknown side

Let's draw a right-angled triangle. Let $\theta$ be one of the acute angles. We are given $\sec \theta = \frac{13}{12}$. We know $\sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent side}}$. Let the Hypotenuse ($AC$) be $\mathbf{13k}$. Let the Adjacent side ($AB$) be $\mathbf{12k}$. We use the Pythagoras theorem. $AC^2 = AB^2 + BC^2$

$(13k)^2 = (12k)^2 + BC^2$

$169k^2 = 144k^2 + BC^2$

$BC^2 = 169k^2 - 144k^2$

$BC^2 = 25k^2$

$BC = \sqrt{25k^2}$

$\boxed{BC = 5k}$

Step 2 — Calculate other ratios

Now we have all three sides. We can find the other trigonometric ratios. Let's find $\sin \theta$. $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC}$

$= \frac{5k}{13k}$

$= \frac{5}{13}$

Let's find $\cos \theta$. $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC}$

$= \frac{12k}{13k}$

$= \frac{12}{13}$

Let's find $\tan \theta$. $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB}$

$= \frac{5k}{12k}$

$= \frac{5}{12}$

Let's find $\csc \theta$. $\csc \theta = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{AC}{BC}$

$= \frac{13k}{5k}$

$= \frac{13}{5}$

Let's find $\cot \theta$. $\cot \theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{AB}{BC}$

$= \frac{12k}{5k}$

$= \frac{12}{5}$

Answer

(i) $\sin \theta = \frac{5}{13}$ (ii) $\cos \theta = \frac{12}{13}$ (iii) $\tan \theta = \frac{5}{12}$ (iv) $\csc \theta = \frac{13}{5}$ (v) $\cot \theta = \frac{12}{5}$