OP Malhotra Class 9 Maths Solutions Chapter 19 Trigonometrical Ratios Exercise 19 (A)

S Chand Class 9 ICSE Maths Solutions Chapter 19 Trigonometrical Ratios Ex 19(A)

Question 1. Complete the following:
Answer: To complete the table, we need to find the trigonometric ratios (sine, cosine, tangent, cosecant, secant, cotangent) for angles A and B in three different right-angled triangles. We use the basic definitions:
\( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \)
\( \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} \)
\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \)
The reciprocal ratios are:
\( \cot \theta = \frac{1}{\tan \theta} = \frac{\text{Base}}{\text{Perpendicular}} \)
\( \sec \theta = \frac{1}{\cos \theta} = \frac{\text{Hypotenuse}}{\text{Base}} \)
\( \csc \theta = \frac{1}{\sin \theta} = \frac{\text{Hypotenuse}}{\text{Perpendicular}} \)
We apply these formulas to each triangle. For the first triangle, sides are 3, 4, 5. For the second, 9, 12, 15 (which is 3 times 3, 4, 5). For the third, 15, 20, 25 (which is 5 times 3, 4, 5).

🎯 Exam Tip: Always label the sides (Perpendicular, Base, Hypotenuse) correctly relative to the angle you are considering. Remember that the hypotenuse is always the longest side, opposite the 90-degree angle.

Question 2. From the figure, find the value of \( \sin \theta, \cos \theta, \tan \theta, \sin^2 \theta, \cos^2 \theta \) and \( \tan^2 \theta \).
Answer: In the given right-angled triangle, we have sides \( AB = 5 \), \( BC = 12 \), and the hypotenuse \( AC = 13 \). The angle \( \theta \) is at vertex C (i.e., \( \angle ACB = \theta \)).
For angle \( \theta \):
Perpendicular (opposite side) \( = AB = 5 \)
Base (adjacent side) \( = BC = 12 \)
Hypotenuse \( = AC = 13 \)
Now, we calculate the required trigonometric ratios:
\( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{5}{13} \)
\( \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{12}{13} \)
\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC} = \frac{5}{12} \)
Next, we find the squared values:
\( \sin^2 \theta = \left(\frac{5}{13}\right)^2 = \frac{5^2}{13^2} = \frac{25}{169} \)
\( \cos^2 \theta = \left(\frac{12}{13}\right)^2 = \frac{12^2}{13^2} = \frac{144}{169} \)
\( \tan^2 \theta = \left(\frac{5}{12}\right)^2 = \frac{5^2}{12^2} = \frac{25}{144} \)
In simple words: First, identify the sides of the triangle relative to angle theta (opposite, adjacent, hypotenuse). Then, use the SOH CAH TOA rule to find sine, cosine, and tangent. To find the squared values, simply multiply the fraction by itself.

🎯 Exam Tip: Always remember the Pythagorean triplet (5, 12, 13) for right-angled triangles; it's a common one that can save time in calculations.

Question 3. From the figure, figure, find the value of \( \tan \theta, \tan^2 \theta, \tan^3 \theta, \sin^2 \theta \) and \( \cos^3 \theta \).
Answer: In the given right-angled triangle, \( \angle C = 90^\circ \) and \( \angle ABC = \theta \). The sides are \( AB = 5 \), \( BC = 3 \), and \( AC = 4 \).
For angle \( \theta \) (at vertex B):
Perpendicular (opposite side) \( = AC = 4 \)
Base (adjacent side) \( = BC = 3 \)
Hypotenuse \( = AB = 5 \)
Now, we calculate the required trigonometric ratios and their powers:
\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AC}{BC} = \frac{4}{3} \)
\( \tan^2 \theta = \left(\frac{4}{3}\right)^2 = \frac{4^2}{3^2} = \frac{16}{9} \)
\( \tan^3 \theta = \left(\frac{4}{3}\right)^3 = \frac{4^3}{3^3} = \frac{64}{27} \)
To find \( \sin^2 \theta \) and \( \cos^3 \theta \), we first need \( \sin \theta \) and \( \cos \theta \):
\( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AC}{AB} = \frac{4}{5} \)
\( \sin^2 \theta = \left(\frac{4}{5}\right)^2 = \frac{4^2}{5^2} = \frac{16}{25} \)
\( \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{BC}{AB} = \frac{3}{5} \)
\( \cos^3 \theta = \left(\frac{3}{5}\right)^3 = \frac{3^3}{5^3} = \frac{27}{125} \)
In simple words: Identify the sides of the right triangle based on angle theta. Calculate the tangent, sine, and cosine ratios using the sides. Then, raise these fractions to the power asked for in the question. Remember, squaring means multiplying by itself once, and cubing means multiplying by itself twice.

🎯 Exam Tip: Pay close attention to which angle is designated as \( \theta \) and make sure your perpendicular and base sides are correctly identified relative to that angle.

Question 4. In the figure, \( \angle OMP, \angle ORQ, \angle OQM \) are right angles. Write the values of the following t-ratios:
(a) \( \sin RQM \)
(b) \( \sin QMP \)
(c) \( \sin OQR \)
(d) \( \cos QMP \)
(e) \( \tan RQM \)
(f) \( \cot MOP \)
(g) \( \sec ROQ \)
Answer: We need to determine the trigonometric ratios for the specified angles by looking at the given figure and identifying the sides of the relevant right-angled triangles. The problem states that \( \angle OMP, \angle ORQ, \angle OQM \) are right angles, which helps us identify the hypotenuse for each triangle.
(a) For \( \sin RQM \): In \( \triangle RQM \), \( \angle RQM \) is the angle. The side opposite to it is \( RM \), and the hypotenuse is \( QM \).
\( \sin RQM = \frac{RM}{QM} \)
(b) For \( \sin QMP \): In \( \triangle QMP \), \( \angle QMP \) is the angle. The side opposite to it is \( QP \), and the hypotenuse is \( PM \).
\( \sin QMP = \frac{QP}{PM} \)
(c) For \( \sin OQR \): In \( \triangle OQR \), \( \angle OQR \) is the angle. The side opposite to it is \( OR \), and the hypotenuse is \( OQ \).
\( \sin OQR = \frac{OR}{OQ} \)
(d) For \( \cos QMP \): In \( \triangle QMP \), \( \angle QMP \) is the angle. The side adjacent to it is \( QM \), and the hypotenuse is \( PM \).
\( \cos QMP = \frac{QM}{PM} \)
(e) For \( \tan RQM \): In \( \triangle RQM \), \( \angle RQM \) is the angle. The side opposite to it is \( RM \), and the side adjacent to it is \( QR \).
\( \tan RQM = \frac{RM}{QR} \)
(f) For \( \cot MOP \): In \( \triangle MOP \), \( \angle MOP \) is the angle. The side adjacent to it is \( OM \), and the side opposite to it is \( MP \).
\( \cot MOP = \frac{OM}{MP} \)
(g) For \( \sec ROQ \): In \( \triangle ROQ \), \( \angle ROQ \) is the angle. The hypotenuse is \( OQ \), and the side adjacent to it is \( OR \).
\( \sec ROQ = \frac{OQ}{OR} \)
In simple words: For each part, first find the specific right-angled triangle that contains the given angle. Then, for that angle, identify the opposite side, adjacent side, and hypotenuse. Finally, apply the correct trigonometric formula (SOH CAH TOA and their reciprocals) to write down the ratio of the sides.

🎯 Exam Tip: When dealing with complex figures, always clearly identify the specific right-angled triangle and the angle in question before determining the opposite, adjacent, and hypotenuse sides.