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Q3:
If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$.

Solution :

Given: In a right-angled triangle $ABC$, where $\angle B = 90^\circ$, we are given that $\sin A = \frac{3}{4}$.

To find: The values of $\cos A$ and $\tan A$.

A B C Adjacent (AB) Opposite (BC) Hypotenuse (AC)

Step 1: Understanding the Trigonometric Ratio
By definition, in a right-angled triangle, the sine of an angle is the ratio of the side opposite to the angle to the hypotenuse. $\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC}$.
Given $\sin A = \frac{3}{4}$, we can assume $BC = 3k$ and $AC = 4k$, where $k$ is a positive constant.

Step 2: Applying the Pythagoras Theorem
In $\triangle ABC$, by the Pythagoras Theorem: $AC^2 = AB^2 + BC^2$
Substituting the known values: $(4k)^2 = AB^2 + (3k)^2$
$16k^2 = AB^2 + 9k^2$
$AB^2 = 16k^2 - 9k^2$
$AB^2 = 7k^2$
$AB = \sqrt{7k^2} = k\sqrt{7}$

Step 3: Calculating $\cos A$
By definition, $\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC}$.
Substituting the values: $\cos A = \frac{k\sqrt{7}}{4k}$
$\cos A = \frac{\sqrt{7}}{4}$

Step 4: Calculating $\tan A$
By definition, $\tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB}$.
Substituting the values: $\tan A = \frac{3k}{k\sqrt{7}}$
$\tan A = \frac{3}{\sqrt{7}}$
To rationalize the denominator: $\tan A = \frac{3}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{3\sqrt{7}}{7}$

Final Answer: $\cos A = \frac{\sqrt{7}}{4}$ and $\tan A = \frac{3\sqrt{7}}{7}$


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