Given that Cos A = $\frac{12}{13}$ for 0 ≤ A ≤ 90º, find Tan A.

1. Analyze the question

We are given the value of $\cos A$ in a right-angled triangle (since $0^\circ \le A \le 90^\circ$, it lies in the first quadrant). We need to find the value of $\tan A$.

2. Explain the concept

In trigonometry, for a right-angled triangle:

• $\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}}$
• $\tan A = \frac{\text{Opposite}}{\text{Adjacent}}$

We can use the Pythagorean theorem ($a^2 + b^2 = c^2$) to find the missing side (the Opposite side) when we know the Adjacent and the Hypotenuse.

3. Show the solution

• Step 1: Identify the given values. $\cos A = \frac{12}{13} = \frac{\text{Adjacent}}{\text{Hypotenuse}}$ So, Adjacent = 12 and Hypotenuse = 13.

• Step 2: Find the Opposite side using Pythagoras theorem. $\text{Opposite}^2 + \text{Adjacent}^2 = \text{Hypotenuse}^2$ $\text{Opposite}^2 + 12^2 = 13^2$ $\text{Opposite}^2 + 144 = 169$ $\text{Opposite}^2 = 169 - 144$ $\text{Opposite}^2 = 25$ $\text{Opposite} = \sqrt{25} = 5$

• Step 3: Calculate $\tan A$. $\tan A = \frac{\text{Opposite}}{\text{Adjacent}}$ $\tan A = \frac{5}{12}$

4. Identify the answer

Comparing our result to the given options, the correct value is $\frac{5}{12}$. This matches option A.

5. Tip

Always remember the "SOH CAH TOA" mnemonic:

• SOH: Sine = Opposite / Hypotenuse
• CAH: Cosine = Adjacent / Hypotenuse
• TOA: Tangent = Opposite / Adjacent

Also, keep in mind the common Pythagorean triples like (5, 12, 13) and (3, 4, 5). Recognizing these can save you valuable time during the JAMB exam!