Question

Given \(\tan A = \frac{4}{3}\), then \(\cos A\) is :

• A. \(\frac{4}{5}\)
• B. \(\frac{3}{5}\)
• C. \(\frac{5}{3}\)
• D. \(\frac{5}{4}\)

Hint:

Remember the standard Pythagorean triplet \((3, 4, 5)\).
Since \(\tan A = \frac{4}{3} = \frac{\text{Opposite}}{\text{Adjacent}}\), the hypotenuse must be \(5\).
Therefore, \(\cos A\), which is \(\frac{\text{Adjacent}}{\text{Hypotenuse}}\), is immediately \(\frac{3}{5}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question is from the topic of Trigonometry.
We are given one trigonometric ratio (\(\tan A\)) and we need to find another trigonometric ratio (\(\cos A\)).

Step 2: Key Formula or Approach:
For a right-angled triangle:
\[ \tan A = \frac{\text{Opposite (Perpendicular)}}{\text{Adjacent (Base)}} \]
\[ \cos A = \frac{\text{Adjacent (Base)}}{\text{Hypotenuse}} \]
By Pythagoras Theorem, we can determine the Hypotenuse (\(H\)) if we know the Perpendicular (\(P\)) and Base (\(B\)):
\[ H^2 = P^2 + B^2 \]

Step 3: Detailed Explanation:
We are given:
\[ \tan A = \frac{4}{3} \]
Thus, we can let:
\[ \text{Perpendicular } (P) = 4k \]
\[ \text{Base } (B) = 3k \]
where \(k\) is a positive constant.
Using the Pythagorean Theorem to find the hypotenuse (\(H\)):
\[ H = \sqrt{P^2 + B^2} \]
\[ H = \sqrt{(4k)^2 + (3k)^2} \]
\[ H = \sqrt{16k^2 + 9k^2} \]
\[ H = \sqrt{25k^2} \]
\[ H = 5k \]
Now, we calculate \(\cos A\):
\[ \cos A = \frac{\text{Base}}{\text{Hypotenuse}} \]
\[ \cos A = \frac{3k}{5k} \]
Cancel \(k\):
\[ \cos A = \frac{3}{5} \]

Step 4: Final Answer:
The value of \(\cos A\) is \(\frac{3}{5}\).