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Introduction to Trigonometry and Its Applications
NCERT Exemplar Ch 08
CBSE_NCERT_EXEMPLAR_CH08
Grade 10
Question:
A pole $6\text{ m}$ high casts a shadow $2\sqrt{3}\text{ m}$ long on the ground, then the sun's elevation is:
$60^\circ$
$45^\circ$
$30^\circ$
$90^\circ$
$60^\circ$
$45^\circ$
$30^\circ$
$90^\circ$
Step-by-Step Solution
Key Concept: $\tan \theta = \dfrac{\text{Height}}{\text{Shadow}} = \dfrac{6}{2\sqrt{3}} = \sqrt{3}$.
Stepwise Solution:
$\tan \theta = \dfrac{6}{2\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3}$. [0.5 Mark]
$\tan \theta = \sqrt{3} \Rightarrow \theta = 60^\circ$. [0.5 Mark]
Marking Scheme:
• Calculating $\tan \theta = \sqrt{3}$: 0.5 Mark
• Finding $\theta = 60^\circ$: 0.5 Mark
$\tan \theta = \dfrac{6}{2\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3}$. [0.5 Mark]
$\tan \theta = \sqrt{3} \Rightarrow \theta = 60^\circ$. [0.5 Mark]
Marking Scheme:
• Calculating $\tan \theta = \sqrt{3}$: 0.5 Mark
• Finding $\theta = 60^\circ$: 0.5 Mark
Correct Answer: $60^\circ$
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