CBSE Class 10 Mathematics Trigonometry Assignment Set 12

Question. In the adjoining figure where \( \triangle ABC \) is right-angled at \( B \) with \( AC = 5\text{ cm} \), \( AB = 4\text{ cm} \), and \( BC = 3\text{ cm} \), with \( \angle ACB = \alpha \) and \( \angle CAB = \beta \), find \( \sin \alpha \).
Answer: \( \frac{4}{5} \)

Question. In the adjoining figure where \( \triangle ABC \) is right-angled at \( B \) with \( AC = 5\text{ cm} \), \( AB = 4\text{ cm} \), and \( BC = 3\text{ cm} \), with \( \angle ACB = \alpha \) and \( \angle CAB = \beta \), find \( \sin \beta \).
Answer: \( \frac{3}{5} \)

Question. In the adjoining figure where \( \triangle ABC \) is right-angled at \( B \) with \( AC = 5\text{ cm} \), \( AB = 4\text{ cm} \), and \( BC = 3\text{ cm} \), with \( \angle ACB = \alpha \) and \( \angle CAB = \beta \), find \( \tan \alpha \cdot \tan \beta \).
Answer: \( 1 \)

Question. In the adjoining figure where \( \triangle ABC \) is right-angled at \( B \) with \( AC = 5\text{ cm} \), \( AB = 4\text{ cm} \), and \( BC = 3\text{ cm} \), with \( \angle ACB = \alpha \) and \( \angle CAB = \beta \), find \( \sec \alpha \div \sec \beta \).
Answer: \( \frac{4}{3} \)

Question. In the adjoining figure where \( \triangle ABC \) is right-angled at \( B \) with \( AC = 5\text{ cm} \), \( AB = 4\text{ cm} \), and \( BC = 3\text{ cm} \), with \( \angle ACB = \alpha \) and \( \angle CAB = \beta \), find \( \sin^2 \alpha + \cos^2 \alpha \).
Answer: \( 1 \)

Question. Express \( \tan A \) in terms of \( \sin A \).
Answer: \( \frac{\sin A}{\sqrt{1 - \sin^2 A}} \)

Question. Express \( \sec A \) in terms of \( \tan A \).
Answer: \( \sqrt{1 + \tan^2 A} \)

Question. Express \( \text{cosec } A \) in terms of \( \cot A \).
Answer: \( \sqrt{1 + \cot^2 A} \)

Question. Without using triangle method, find \( \sin A \) if \( \cos A = \frac{5}{13} \).
Answer: \( \frac{12}{13} \)

Question. Without using triangle method, find \( \tan A \) if \( \sec A = \frac{13}{5} \).
Answer: \( \frac{12}{5} \)

Question. Without using triangle method, find \( \text{cosec } A \) if \( \cot A = \frac{5}{12} \).
Answer: \( \frac{13}{12} \)

Question. Fill in the blank: \( \frac{1 + \sin A}{\cos A} = \frac{\text{______}}{1 - \sin A} \)
Answer: \( \cos A \)

Question. Find the value of \( \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \cdot \tan 30^\circ} \).
Answer: \( 2 + \sqrt{3} \)

Question. If \( 2 \cos A = \sqrt{3} \), then find \( \angle A \).
Answer: \( 30^\circ \)

Question. If \( \sin \theta = \sqrt{3} \cos \theta \), then find \( \angle \theta \).
Answer: \( 60^\circ \)

Question. If \( x = a \sin \theta \) and \( y = a \cos \theta \), find the relation between \( x \), \( y \), and \( a \).
Answer: \( x^2 + y^2 = a^2 \)

Question. If \( \angle A + \angle B + \angle C = 180^\circ \), then find the value of \( \sin\left(\frac{\angle A + \angle B}{2}\right) \).
Answer: \( \cos \frac{C}{2} \)

Question. If \( \tan A = \cot B \), then find \( A + B \).
Answer: \( 90^\circ \)

Question. If \( \sin 3A = \cos(A - 30^\circ) \), then find \( \angle A \).
Answer: \( 30^\circ \)