Grade 12 Examination Question - Trigonometry

This question appeared in 2014 Grade 12 Examination Advance mathematics. 

Finding tan A in a Right-Angled Triangle

Finding tan A in a Right-Angled Triangle

In this post, we will solve for tan A in the given right-angled triangle and express our answer in exact form.

Step 1: Understanding the Given Triangle

We are given a right-angled triangle with:

• Opposite side = 2
• Hypotenuse = 3
• Adjacent side = x (unknown)

Using the Pythagorean Theorem:

$$a^2+b^2=c^2$$

Substituting the known values:

$$x^2+2^2=3^2$$

$$x^2+4=9$$

$$x^2=5$$

$$x=\sqrt{5}$$

Step 2: Applying the Tangent Formula

Tangent is defined as:

$$\tan A=\frac{\text{opposite}}{\text{adjacent}}$$

Substituting the values:

$$\tan A=\frac{2}{\sqrt{5}}$$

Step 3: Rationalizing the Denominator

To eliminate the square root in the denominator, we multiply by $\frac{\sqrt{5}}{\sqrt{5}}$:

$$\tan A=\frac{2}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}$$

$$\tan A=\frac{2\sqrt{5}}{5}$$

Final Answer

$$\tan A=\frac{2\sqrt{5}}{5}$$

Thus, the exact value of tan A is $\frac{2\sqrt{5}}{5}$.

Conclusion

By following these steps, we were able to find the exact value of tan A in the given right-angled triangle. Remember to always rationalize the denominator for a clean and simplified final answer. 

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