Introduction to Trigonometry and Its Applications
NCERT Exemplar Ch 08
CBSE_NCERT_EXEMPLAR_CH08
Grade 10
Question:
If $\cos A = \dfrac{4}{5}$, then the value of $\tan A$ is:
$\dfrac{3}{5}$
$\dfrac{3}{4}$
$\dfrac{4}{3}$
$\dfrac{5}{3}$
Step-by-Step Solution
Key Concept: $\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - 16/25} = 3/5$. $\tan A = \dfrac{\sin A}{\cos A}$.
Stepwise Solution:
$\sin A = \sqrt{1 - (4/5)^2} = \sqrt{9/25} = \dfrac{3}{5}$. [0.5 Mark]
$\tan A = \dfrac{3/5}{4/5} = \dfrac{3}{4}$. [0.5 Mark]
Marking Scheme:
• Finding $\sin A = 3/5$: 0.5 Mark
• Calculating $\tan A = 3/4$: 0.5 Mark
Correct Answer: $\dfrac{3}{4}$
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