Samacheer Kalvi Class 9 Maths Solutions Chapter 6 Trigonometry Exercise 6.4

Question 1. Find the value of the following:
(i) \( \sin 49^\circ \)
(ii) \( \cos 74^\circ 39' \)
(iii) \( \tan 54^\circ 26' \)
(iv) \( \sin 21^\circ 21' \)
(v) \( \cos 33^\circ 53' \)
(vi) \( \tan 70^\circ 17' \)
Answer:
(i) To find the value of \( \sin 49^\circ \), we look up \( 49^\circ \) in the sine trigonometric table. The sine value for \( 49^\circ \) is \( 0.7547 \).
(ii) To find \( \cos 74^\circ 39' \), we use the cosine table. First, find \( \cos 74^\circ 36' = 0.2656 \). Then, locate the mean difference for \( 3' \), which is 8. For cosine values, we subtract the mean difference. So, \( \cos 74^\circ 39' = 0.2656 - 0.0008 = 0.2648 \).
(iii) To find \( \tan 54^\circ 26' \), we use the tangent table. First, find \( \tan 54^\circ 24' = 1.3968 \). Then, locate the mean difference for \( 2' \), which is 17. For tangent values, we add the mean difference. So, \( \tan 54^\circ 26' = 1.3968 + 0.0017 = 1.3985 \).
(iv) To find \( \sin 21^\circ 21' \), we use the sine table. First, find \( \sin 21^\circ 18' = 0.3633 \). Then, locate the mean difference for \( 3' \), which is 8. For sine values, we add the mean difference. So, \( \sin 21^\circ 21' = 0.3633 + 0.0008 = 0.3641 \).
(v) To find \( \cos 33^\circ 53' \), we use the cosine table. First, find \( \cos 33^\circ 48' = 0.8310 \). Then, locate the mean difference for \( 5' \), which is 8. For cosine values, we subtract the mean difference. So, \( \cos 33^\circ 53' = 0.8310 - 0.0008 = 0.8302 \).
(vi) To find \( \tan 70^\circ 17' \), we use the tangent table. First, find \( \tan 70^\circ 12' = 2.7776 \). Then, locate the mean difference for \( 5' \), which is 131. For tangent values, we add the mean difference. So, \( \tan 70^\circ 17' = 2.7776 + 0.0131 = 2.7907 \).
In simple words: These problems involve looking up values in trigonometric tables. For sine and tangent, you usually add the mean difference. For cosine, you subtract the mean difference. This helps to get values for angles that are not exactly listed in the main table.

🎯 Exam Tip: Always remember to add mean differences for sine and tangent, but subtract them for cosine when using trigonometric tables. This rule is crucial for accuracy.

Question 2. Find the value of \( \theta \) if
(i) \( \sin \theta = 0.9975 \)
(ii) \( \cos \theta = 0.6763 \)
(iii) \( \tan \theta = 0.0720 \)
(iv) \( \cos \theta = 0.0410 \)
(v) \( \tan \theta = 7.5958 \)
Answer:
(i) To find \( \theta \) when \( \sin \theta = 0.9975 \), we check the sine table. We find that \( \sin 85^\circ 54' = 0.9974 \). The difference needed is \( 0.9975 - 0.9974 = 0.0001 \). This difference corresponds to a mean difference of \( 1' \) in the table. So, \( \theta = 85^\circ 54' + 1' = 85^\circ 55' \). Depending on table precision, \( 85^\circ 56' \) or \( 85^\circ 57' \) might also be found.
(ii) To find \( \theta \) when \( \cos \theta = 0.6763 \), we check the cosine table. We find that \( \cos 47^\circ 24' = 0.6769 \). The difference between \( 0.6769 \) and \( 0.6763 \) is \( 0.0006 \). This difference corresponds to a mean difference of \( 3' \). Since it is cosine, we add the mean difference to the angle, so \( \theta = 47^\circ 24' + 3' = 47^\circ 27' \).
(iii) To find \( \theta \) when \( \tan \theta = 0.0720 \), we check the tangent table. We find that \( \tan 4^\circ 6' = 0.0717 \). The difference needed is \( 0.0720 - 0.0717 = 0.0003 \). This difference corresponds to a mean difference of \( 1' \). So, \( \theta = 4^\circ 6' + 1' = 4^\circ 7' \).
(iv) To find \( \theta \) when \( \cos \theta = 0.0410 \), we check the cosine table. We find that \( \cos 87^\circ 36' = 0.0419 \). The difference between \( 0.0419 \) and \( 0.0410 \) is \( 0.0009 \). This difference corresponds to a mean difference of \( 3' \). For cosine, we add the mean difference to the angle, so \( \theta = 87^\circ 36' + 3' = 87^\circ 39' \).
(v) To find \( \theta \) when \( \tan \theta = 7.5958 \), we check the tangent table directly. We find that \( \tan 82^\circ 30' = 7.5958 \). In this case, the value matches exactly, so there is no mean difference to add. Thus, \( \theta = 82^\circ 30' \).
In simple words: To find an angle from its trigonometric value, you look up the value in the correct table. If the exact value isn't there, you find the closest one and use the mean difference column to adjust the angle. Remember, for cosine, you add the mean difference to the angle when working backwards from the table.

🎯 Exam Tip: When finding angles, pay close attention to whether the mean difference needs to be added or subtracted, depending on the trigonometric function and the table's layout. A common mistake is using the wrong operation.

Question 3. Find the value of the following:
(i) \( \sin 65^\circ 39' + \cos 24^\circ 57' + \tan 10^\circ 10' \)
(ii) \( \tan 70^\circ 58' + \cos 15^\circ 26' - \sin 84^\circ 59' \)
Answer:
(i) First, we find the individual values from the trigonometric tables:
\( \sin 65^\circ 39' = 0.9111 \)
\( \cos 24^\circ 57' = 0.9066 \)
\( \tan 10^\circ 10' = 0.1793 \)
Now, we add these values together as per the expression:
\( 0.9111 + 0.9066 + 0.1793 = 1.9970 \)
So, the value of \( \sin 65^\circ 39' + \cos 24^\circ 57' + \tan 10^\circ 10' \) is \( 1.9970 \).
(ii) First, we find the individual values from the trigonometric tables:
\( \tan 70^\circ 58' = 2.8982 \)
\( \cos 15^\circ 26' = 0.9639 \)
\( \sin 84^\circ 59' = 0.9962 \)
Now, we perform the calculation according to the expression:
\( 2.8982 + 0.9639 - 0.9962 = 3.8621 - 0.9962 = 2.8659 \)
So, the value of \( \tan 70^\circ 58' + \cos 15^\circ 26' - \sin 84^\circ 59' \) is \( 2.8659 \).
In simple words: For these types of problems, you need to find each trigonometric value separately from a table first. Then, you simply perform the addition and subtraction as shown in the problem to get the final answer. It's like doing multiple small calculations before the final sum or difference.

🎯 Exam Tip: Always calculate each trigonometric value separately before combining them. Double-check your table readings and the operations (addition/subtraction) to avoid simple errors.

Question 4. Find the area of a right triangle whose hypotenuse is 10 cm and one of the acute angle is \( 24^\circ 24' \).
Answer: Let the right triangle be ABC, with the right angle at B. Let AC be the hypotenuse, so \( AC = 10 \text{ cm} \). Let one acute angle be \( \angle C = 24^\circ 24' \). To find the area, we need the lengths of the two perpendicular sides, AB and BC.

Using trigonometry in the right triangle ABC:
To find side AB (the side opposite to \( \angle C \)):
\( \sin C = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} \)
\( \sin 24^\circ 24' = \frac{AB}{10} \)
From the trigonometric table, \( \sin 24^\circ 24' \approx 0.4131 \).
So, \( 0.4131 = \frac{AB}{10} \)
\( AB = 0.4131 \times 10 = 4.131 \text{ cm} \).
To find side BC (the side adjacent to \( \angle C \)):
\( \cos C = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} \)
\( \cos 24^\circ 24' = \frac{BC}{10} \)
From the trigonometric table, \( \cos 24^\circ 24' \approx 0.9107 \).
So, \( 0.9107 = \frac{BC}{10} \)
\( BC = 0.9107 \times 10 = 9.107 \text{ cm} \).
Now, we calculate the area of the right triangle:
Area \( = \frac{1}{2} \times \text{base} \times \text{height} \)
Area \( = \frac{1}{2} \times BC \times AB \)
Area \( = \frac{1}{2} \times 9.107 \times 4.131 \)
Area \( = \frac{37.620097}{2} \)
Area \( = 18.81 \text{ cm}^2 \) (rounded to two decimal places). The area calculation for a right-angled triangle is straightforward once its two shorter sides are known.
In simple words: First, use sine and cosine to find the lengths of the two shorter sides of the triangle. The longest side (hypotenuse) and one angle are given. Once you have the two perpendicular sides, multiply them and divide by two to get the triangle's area.

🎯 Exam Tip: When dealing with right triangles, remember SOH CAH TOA to correctly identify which trigonometric ratio (sine, cosine, or tangent) to use for finding unknown sides or angles. Always double-check calculations.

Question 5. Find the angle made by a ladder of length 5m with the ground, if one of its end is 4m away from the wall and the other end is on the wall.
Answer: Let AC be the ladder, with its length \( AC = 5 \text{ m} \). Let BC be the distance from the foot of the ladder to the base of the wall, so \( BC = 4 \text{ m} \). The other end of the ladder, A, rests against the wall. We need to find the angle \( \theta \) that the ladder makes with the ground, which is \( \angle C \).

In the right triangle ABC (which is right-angled at B, where the wall meets the ground):
We know the adjacent side (BC = 4 m) and the hypotenuse (AC = 5 m). The cosine function relates these two sides to the angle.
\( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} \)
\( \cos \theta = \frac{4}{5} \)
\( \cos \theta = 0.8 \)
Now, we find the angle \( \theta \) whose cosine is \( 0.8 \). From trigonometric tables, we can find that \( \cos 36^\circ 52' \approx 0.8000 \).
Therefore, \( \theta = 36^\circ 52' \). This means the ladder makes an angle of \( 36^\circ 52' \) with the ground. This problem is a direct application of trigonometry to a real-life scenario.
In simple words: Imagine a right triangle formed by the wall, the ground, and the ladder. The ladder is the longest side (hypotenuse), and the distance from the wall is the base (adjacent side). Use the cosine rule (adjacent side divided by hypotenuse) to find the angle the ladder makes with the ground.

🎯 Exam Tip: When visualizing real-world problems like ladders against walls, always draw a right-angled triangle to correctly identify the hypotenuse, opposite, and adjacent sides relative to the angle you need to find. This prevents errors in applying trigonometric ratios.

Question 6. In the given figure, HT shows the height of a tree standing vertically. From a point P, the angle of elevation of the top of the tree (that is \( \angle P \)) measures \( 42^\circ \) and the distance to the tree is 60 metres. Find the height of the tree.
Answer: Let HT be the height of the tree, which we will call 'x'. The tree stands straight up, so it forms a right angle with the ground at point T. From a point P on the ground, the angle of elevation to the top of the tree (H) is \( \angle P = 42^\circ \). The distance from point P to the base of the tree (T) is given as \( PT = 60 \text{ metres} \).

In the right triangle HTP (right-angled at T):
We have the angle of elevation \( \angle P = 42^\circ \), the adjacent side \( PT = 60 \text{ m} \), and we need to find the opposite side HT (the height of the tree). The tangent function connects the opposite and adjacent sides relative to the angle.
\( \tan P = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{HT}{PT} \)
\( \tan 42^\circ = \frac{x}{60} \)
From the trigonometric table, \( \tan 42^\circ \approx 0.9004 \).
So, \( 0.9004 = \frac{x}{60} \)
\( x = 0.9004 \times 60 \)
\( x = 54.024 \)
Therefore, the height of the tree is approximately \( 54.02 \text{ m} \) (rounded to two decimal places). This example shows how trigonometry is very useful for finding heights of tall objects without directly measuring them.
In simple words: Think of the tree, the ground, and the line of sight to the treetop as a right triangle. The distance to the tree is the base (adjacent side), and the height of the tree is the vertical side (opposite side). Use the tangent function (opposite divided by adjacent) to find the height.

🎯 Exam Tip: For angle of elevation problems, always draw a clear diagram and identify which trigonometric ratio (sine, cosine, or tangent) connects the given angle, the known side, and the unknown side. This helps in choosing the correct formula.