CBSE Class 10 Mathematics Basic sample paper 2026-27: question 29

Hints

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Step-by-step solution

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1. By Pythagoras theorem, $AC=\sqrt{A{B}^2-B{C}^2}=\sqrt{29^2-21^2}=\sqrt{400}=20$ units

   ½ mark

2. For $\angle B=\theta$: $\sin \theta =\frac{AC}{AB}=\frac{20}{29}$, $\cos \theta =\frac{BC}{AB}=\frac{21}{29}$, $\tan \theta =\frac{20}{21}$

   ½ mark

3. (i) $1+{\tan }^2\theta =1+\frac{400}{441}=\frac{841}{441}$

   1 mark

4. (ii) ${\cos }^2\theta -{\sin }^2\theta =\frac{441}{841}-\frac{400}{841}=\frac{41}{841}$

   1 mark

Answer: (i) $\frac{841}{441}$; (ii) $\frac{41}{841}$

Revise the chapter

• Chapter: Introduction to Trigonometry (notes, formulas, where marks are lost and every past question)
• Topic: Relationships Between Trigonometric Ratios