If sin A = \(\dfrac{3}{4}\), calculate cos A and tan A.
cos A = \(\dfrac{\sqrt{7}}{4}\), tan A = \(\dfrac{3}{\sqrt{7}}\)
Given: \(\sin A = \dfrac{3}{4}\).
To find: \(\cos A\) and \(\tan A\).
Step 1: Understand what \(\sin A\) means
In a right-angled triangle (taking \(A\) as an acute angle),
\(\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}}\).
So we can assume:
Opposite side = \(3\) units and Hypotenuse = \(4\) units.
Student Note: This is a standard trick: when a ratio like \(\dfrac{3}{4}\) is given, we choose sides in the same ratio (3 and 4). The final trigonometric values remain the same.
Step 2: Find the adjacent side using Pythagoras theorem
Using \(\text{(Hypotenuse)}^2 = \text{(Opposite)}^2 + \text{(Adjacent)}^2\):
\(4^2 = 3^2 + (\text{Adjacent})^2\)
\(16 = 9 + (\text{Adjacent})^2\)
\((\text{Adjacent})^2 = 16 - 9 = 7\)
So, \(\text{Adjacent} = \sqrt{7}\) units.
Step 3: Calculate \(\cos A\)
\(\cos A = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{\sqrt{7}}{4}\).
Step 4: Calculate \(\tan A\)
\(\tan A = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{3}{\sqrt{7}}\).
Student Note (Optional): Sometimes we rationalize the denominator:
\(\tan A = \dfrac{3}{\sqrt{7}} \times \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{3\sqrt{7}}{7}\). Both forms are correct unless your book specifically asks for rationalized form.
Quick Check: Since \(\sin^2 A + \cos^2 A = 1\),
\(\left(\dfrac{3}{4}\right)^2 + \left(\dfrac{\sqrt{7}}{4}\right)^2 = \dfrac{9}{16} + \dfrac{7}{16} = 1\). Correct.