CBSE Class 10 Mathematics Trigonometry Assignment Set 22

Trigonometric Ratios of an Angle (Exercise I)

Question 1. In \( \triangle ABC \), right-angled at B, \( AB = 24 \) cm, \( BC = 7 \) cm. Determine: (i) \( \sin A,\ \cos A \) (ii) \( \sin C,\ \cos C \).

Answer:
By Pythagoras theorem, \( AC^2 = AB^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625 \Rightarrow AC = 25 \) cm.
(i) For \( \angle A \): Perpendicular \( = BC = 7 \), Base \( = AB = 24 \). \( \sin A = \frac{BC}{AC} = \frac{7}{25} \) and \( \cos A = \frac{AB}{AC} = \frac{24}{25} \).
(ii) For \( \angle C \): Perpendicular \( = AB = 24 \), Base \( = BC = 7 \). \( \sin C = \frac{AB}{AC} = \frac{24}{25} \) and \( \cos C = \frac{BC}{AC} = \frac{7}{25} \).
In simple words: The side opposite the angle is the perpendicular, so the sine of A and the sine of C swap the two short sides.

Teacher's Note:
1. Draw the right triangle, mark the right angle and name the perpendicular, base and hypotenuse with respect to the angle asked.
2. Always find the third side with the Pythagoras theorem before writing any ratio.

Question 2. In the given figure, \( \triangle PQR \) is right-angled at Q with \( PQ = 12 \) cm and \( PR = 13 \) cm. Find \( \tan P - \cot R \).

Answer:
By Pythagoras theorem in \( \triangle PQR \): \( PR^2 = PQ^2 + QR^2 \Rightarrow 169 = 144 + QR^2 \Rightarrow QR^2 = 25 \Rightarrow QR = 5 \) cm.
\( \tan P = \frac{\text{side opposite to } \angle P}{\text{side adjacent to } \angle P} = \frac{QR}{PQ} = \frac{5}{12} \).
\( \cot R = \frac{\text{side adjacent to } \angle R}{\text{side opposite to } \angle R} = \frac{QR}{PQ} = \frac{5}{12} \).
\( \tan P - \cot R = \frac{5}{12} - \frac{5}{12} = 0 \).
In simple words: Both ratios compare the same two sides, so they are equal and their difference is zero.

Teacher's Note:
1. For angle R the side opposite is PQ and the side adjacent is QR.
2. Check each ratio against the angle it belongs to before subtracting.

Question 3. If \( \sin A = \frac{3}{4} \), calculate \( \cos A \) and \( \tan A \).
Answer:
\( \sin A = \frac{BC}{AC} = \frac{3}{4} \). Let \( BC = 3k \) and \( AC = 4k \), where k is a positive number.
By Pythagoras theorem, \( AB^2 = AC^2 - BC^2 = 16k^2 - 9k^2 = 7k^2 \Rightarrow AB = \sqrt{7}\,k \).
\( \cos A = \frac{AB}{AC} = \frac{\sqrt{7}k}{4k} = \frac{\sqrt{7}}{4} \) and \( \tan A = \frac{BC}{AB} = \frac{3k}{\sqrt{7}k} = \frac{3}{\sqrt{7}} \).
In simple words: Take the sides as 3k and 4k, find the third side and write the required ratios.

Teacher's Note:
1. Draw the right triangle, mark the right angle and name the perpendicular, base and hypotenuse with respect to the angle asked.
2. Always find the third side with the Pythagoras theorem before writing any ratio.

Question 4. Given \( 15 \cot A = 8 \), find \( \sin A \) and \( \sec A \).
Answer:
\( \cot A = \frac{8}{15} = \frac{AB}{BC} \). Let \( AB = 8k \) and \( BC = 15k \).
\( AC^2 = AB^2 + BC^2 = 64k^2 + 225k^2 = 289k^2 \Rightarrow AC = 17k \).
\( \sin A = \frac{BC}{AC} = \frac{15k}{17k} = \frac{15}{17} \) and \( \sec A = \frac{AC}{AB} = \frac{17k}{8k} = \frac{17}{8} \).
In simple words: Cotangent is base over perpendicular, so 8 is the base and 15 is the perpendicular.

Teacher's Note:
1. Draw the right triangle, mark the right angle and name the perpendicular, base and hypotenuse with respect to the angle asked.
2. Always find the third side with the Pythagoras theorem before writing any ratio.

Question 5. Given \( \sec \theta = \frac{13}{12} \), calculate all other trigonometric ratios.
Answer:
\( \sec \theta = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{AC}{AB} = \frac{13}{12} \). Let \( AC = 13k \) and \( AB = 12k \).
\( BC^2 = AC^2 - AB^2 = 169k^2 - 144k^2 = 25k^2 \Rightarrow BC = 5k \).
\( \sin \theta = \frac{BC}{AC} = \frac{5}{13} \), \( \cos \theta = \frac{AB}{AC} = \frac{12}{13} \), \( \tan \theta = \frac{BC}{AB} = \frac{5}{12} \), \( \operatorname{cosec} \theta = \frac{AC}{BC} = \frac{13}{5} \), \( \cot \theta = \frac{AB}{BC} = \frac{12}{5} \).
In simple words: Once the three sides 5, 12 and 13 are known, every ratio is read straight from the triangle.

Teacher's Note:
1. Draw the right triangle, mark the right angle and name the perpendicular, base and hypotenuse with respect to the angle asked.
2. Always find the third side with the Pythagoras theorem before writing any ratio.

Question 6. If \( \angle A \) and \( \angle B \) are acute angles such that \( \cos A = \cos B \), then show that \( \angle A = \angle B \).
Answer:
Draw a right triangle ABC with the right angle at C, so that \( \angle A \) and \( \angle B \) are acute.
\( \cos A = \frac{AC}{AB} \) and \( \cos B = \frac{BC}{AB} \). Since \( \cos A = \cos B \), \( \frac{AC}{AB} = \frac{BC}{AB} \Rightarrow AC = BC \).
Angles opposite to equal sides are equal, therefore \( \angle B = \angle A \). Hence proved.
In simple words: Equal cosines give equal sides next to the right angle, so the triangle is isosceles and the two acute angles are equal.

Teacher's Note:
1. Build the triangle with the two given acute angles and the right angle at the third vertex.
2. Use the property: angles opposite equal sides of a triangle are equal.

Question 7. If \( \cot \theta = \frac{7}{8} \), evaluate: (i) \( \frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)} \) (ii) \( \cot^2 \theta \).
Answer:
\( \cot \theta = \frac{BC}{AB} = \frac{7}{8} \). Let \( BC = 7k \) and \( AB = 8k \). Then \( AC^2 = 64k^2 + 49k^2 = 113k^2 \Rightarrow AC = \sqrt{113}\,k \).
\( \sin \theta = \frac{AB}{AC} = \frac{8}{\sqrt{113}} \) and \( \cos \theta = \frac{BC}{AC} = \frac{7}{\sqrt{113}} \).
(i) \( \frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)} = \frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} = \frac{1 - \frac{64}{113}}{1 - \frac{49}{113}} = \frac{49}{64} \).
(ii) \( \cot^2 \theta = \left( \frac{7}{8} \right)^2 = \frac{49}{64} \).
In simple words: The expression in (i) simplifies to cos²θ over sin²θ, which is cot²θ, so both parts give the same number.

Teacher's Note:
1. Notice the identity a² − b² = (a + b)(a − b) in the numerator and denominator.
2. Check: (i) and (ii) must agree because 1 − sin²θ = cos²θ and 1 − cos²θ = sin²θ.

Question 8. If \( 3 \cot A = 4 \), check whether \( \frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A \) or not.
Answer:
\( \cot A = \frac{4}{3} = \frac{AB}{BC} \). Let \( AB = 4k \) and \( BC = 3k \). Then \( AC^2 = 16k^2 + 9k^2 = 25k^2 \Rightarrow AC = 5k \).
\( \tan A = \frac{3}{4},\ \sin A = \frac{3}{5},\ \cos A = \frac{4}{5} \).
LHS \( = \frac{1 - \left( \frac{3}{4} \right)^2}{1 + \left( \frac{3}{4} \right)^2} = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{7}{25} \).
RHS \( = \cos^2 A - \sin^2 A = \left( \frac{4}{5} \right)^2 - \left( \frac{3}{5} \right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} \).
LHS = RHS, so the given relation is true.
In simple words: Work out the three ratios from the sides 3, 4 and 5 and put them on both sides.

Teacher's Note:
1. Find tan A, sin A and cos A first; then evaluate the left and right sides separately.
2. Both sides must be reduced to the same fraction before concluding.

Question 9. In \( \triangle ABC \), right-angled at B, if \( \tan A = \frac{1}{\sqrt{3}} \), find the value of: (i) \( \sin A \cos C + \cos A \sin C \) (ii) \( \cos A \cos C - \sin A \sin C \).
Answer:
\( \tan A = \frac{BC}{AB} = \frac{1}{\sqrt{3}} \). Let \( BC = k \) and \( AB = \sqrt{3}\,k \). Then \( AC^2 = 3k^2 + k^2 = 4k^2 \Rightarrow AC = 2k \).
\( \sin A = \frac{BC}{AC} = \frac{1}{2},\ \cos A = \frac{AB}{AC} = \frac{\sqrt{3}}{2},\ \sin C = \frac{AB}{AC} = \frac{\sqrt{3}}{2},\ \cos C = \frac{BC}{AC} = \frac{1}{2} \).
(i) \( \sin A \cos C + \cos A \sin C = \frac{1}{2} \times \frac{1}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1 \).
(ii) \( \cos A \cos C - \sin A \sin C = \frac{\sqrt{3}}{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{\sqrt{3}}{2} = 0 \).
In simple words: Find the sides in the ratio 1 : √3 : 2, read the ratios of both acute angles and substitute.

Teacher's Note:
1. Remember that for angle C the perpendicular and the base interchange their roles.
2. Compare: sin C = cos A and cos C = sin A because A and C are complementary.

Question 10. In \( \triangle PQR \), right-angled at Q, \( PR + QR = 25 \) cm and \( PQ = 5 \) cm. Determine the values of \( \sin P,\ \cos P \) and \( \tan P \).
Answer:
Let \( QR = x \) cm. Then \( PR = (25 - x) \) cm. By Pythagoras theorem, \( PR^2 = PQ^2 + QR^2 \Rightarrow (25 - x)^2 = 5^2 + x^2 \).
\( 625 - 50x + x^2 = 25 + x^2 \Rightarrow 50x = 600 \Rightarrow x = 12 \). So \( QR = 12 \) cm and \( PR = 13 \) cm.
\( \sin P = \frac{QR}{PR} = \frac{12}{13},\ \cos P = \frac{PQ}{PR} = \frac{5}{13},\ \tan P = \frac{QR}{PQ} = \frac{12}{5} \).
In simple words: Assume one unknown side, form an equation with the Pythagoras theorem and solve it; the sides are 5, 12 and 13.

Teacher's Note:
1. Use the sum given to write the hypotenuse in terms of the unknown side.
2. The x² terms cancel, which leaves a simple linear equation.

Question 11. State whether the following are true or false. Justify your answer. (i) The value of \( \tan A \) is always less than 1. (ii) \( \sec A = \frac{12}{5} \) for some value of angle A. (iii) \( \cos A \) is the abbreviation used for the cosecant of angle A. (iv) \( \cot A \) is the product of cot and A. (v) \( \sin \theta = \frac{4}{3} \) for some angle \( \theta \).
Answer:
(i) False. \( \tan 60^\circ = \sqrt{3} \approx 1.73 \), which is greater than 1.
(ii) True. The value of \( \sec A \) is always greater than or equal to 1, and \( \frac{12}{5} > 1 \).
(iii) False. \( \cos A \) is the abbreviation for the cosine of angle A, not the cosecant.
(iv) False. \( \cot A \) is the cotangent of angle A; it is not the product of cot and A.
(v) False. \( \sin \theta \) can never be greater than 1, and \( \frac{4}{3} > 1 \).
In simple words: Sine and cosine never exceed 1, secant and cosecant are never below 1, and tangent and cotangent can take any value.

Teacher's Note:
1. Remember the range of each ratio: |sin θ| ≤ 1, |cos θ| ≤ 1, |sec θ| ≥ 1 and |cosec θ| ≥ 1.
2. A single counter-example is enough to show a statement is false.