Introduction to Trigonometry
CH08 Question Bank
CBSE_CH08_QUESTION_BANK
Grade 10
Question:
If $\sin A=\dfrac{3}{5}$, find $\cos A$ and $\tan A$ (assuming $A$ is acute).
Step-by-Step Solution
Key Concept: Use the identity $\sin^2A+\cos^2A=1$ to find $\cos A$, then compute $\tan A=\sin A/\cos A$.
$\cos^2A=1-\sin^2A=1-\dfrac{9}{25}=\dfrac{16}{25}\Rightarrow\cos A=\dfrac45$. [1.0 Mark]
$\tan A=\dfrac{\sin A}{\cos A}=\dfrac{3/5}{4/5}=\dfrac34$. [1.0 Mark]
Correct Answer:
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