Test Index

CBSE Class 10 Basic Math 2026 All Sets Solved Paper

© examsnet.com
Question : 5 of 20
Marks: +1, -0
If sin⁡θ=17\sin \theta = \frac{1}{7}, then tan⁡θ\tan \theta is:
Solution:  
Given:
sin⁡ θ=OppositeHypotenuse=17\sin\,\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1}{7}
By Pythagoras:
Base=72−12\text{Base} = \sqrt{7^2 - 1^2}
=49−1= \sqrt{49-1}
=48= \sqrt{48}
=43= 4\sqrt{3}
∴tan⁡ θ=OppositeAdjacent\therefore \tan\,\theta = \frac{\text{Opposite}}{\text{Adjacent}}
=143= \frac{1}{4\sqrt{3}}
© examsnet.com
Go to Question:
Prev Question
Next Question