Samacheer Kalvi Class 9 Maths Solutions Chapter 6 Trigonometry Exercise 6.1

Question 1. From the given figure, find all the trigonometric ratios of angle B.
Answer: From the given right-angled triangle, for angle B:
Opposite side (AC) = 9
Adjacent side (BC) = 40
Hypotenuse (AB) = 41

We can find the trigonometric ratios as follows:
\( \sin B = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{9}{41} \)
\( \cos B = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{40}{41} \)
\( \tan B = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{9}{40} \)
\( \csc B = \frac{\text{Hypotenuse}}{\text{Opposite side}} = \frac{41}{9} \)
\( \sec B = \frac{\text{Hypotenuse}}{\text{Adjacent side}} = \frac{41}{40} \)
\( \cot B = \frac{\text{Adjacent side}}{\text{Opposite side}} = \frac{40}{9} \)
In simple words: We use the lengths of the sides of the triangle (opposite, adjacent, hypotenuse) to calculate sine, cosine, tangent, and their inverse ratios for angle B. Each ratio is a simple fraction of two side lengths.

🎯 Exam Tip: Always correctly identify the opposite, adjacent, and hypotenuse sides relative to the angle in question before calculating the ratios.

Question 2. From the given figure, find the values of
(i) sin B
(ii) sec B
(iii) cot B
(iv) cos C
(v) tan C
(vi) cosec C
Answer:
First, we need to find the missing side lengths in the triangles.
In the right-angled triangle \( \triangle ABD \):
Using Pythagoras theorem: \( AD^2 = AB^2 - BD^2 \)
\( AD^2 = 13^2 - 5^2 \)
\( AD^2 = 169 - 25 \)
\( AD^2 = 144 \)
\( AD = \sqrt{144} \)
\( AD = 12 \)

In the right-angled triangle \( \triangle ADC \):
Using Pythagoras theorem: \( AC^2 = AD^2 + DC^2 \)
\( AC^2 = 12^2 + 16^2 \)
\( AC^2 = 144 + 256 \)
\( AC^2 = 400 \)
\( AC = \sqrt{400} \)
\( AC = 20 \)

Now we can find the trigonometric ratios:
(i) For angle B in \( \triangle ABD \):
Opposite side (AD) = 12
Hypotenuse (AB) = 13
\( \sin B = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{AD}{AB} = \frac{12}{13} \)

(ii) For angle B in \( \triangle ABD \):
Adjacent side (BD) = 5
Hypotenuse (AB) = 13
\( \sec B = \frac{\text{Hypotenuse}}{\text{Adjacent side}} = \frac{AB}{BD} = \frac{13}{5} \)

(iii) For angle B in \( \triangle ABD \):
Opposite side (AD) = 12
Adjacent side (BD) = 5
\( \cot B = \frac{\text{Adjacent side}}{\text{Opposite side}} = \frac{BD}{AD} = \frac{5}{12} \)

(iv) For angle C in \( \triangle ADC \):
Adjacent side (CD) = 16
Hypotenuse (AC) = 20
\( \cos C = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{CD}{AC} = \frac{16}{20} = \frac{4}{5} \)

(v) For angle C in \( \triangle ADC \):
Opposite side (AD) = 12
Adjacent side (CD) = 16
\( \tan C = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{AD}{CD} = \frac{12}{16} = \frac{3}{4} \)

(vi) For angle C in \( \triangle ADC \):
Opposite side (AD) = 12
Hypotenuse (AC) = 20
\( \csc C = \frac{\text{Hypotenuse}}{\text{Opposite side}} = \frac{AC}{AD} = \frac{20}{12} = \frac{5}{3} \)
In simple words: First, we use the Pythagorean theorem to find the lengths of the unknown sides in both triangles. Then, we use these lengths to calculate the sine, secant, cotangent, cosine, and tangent ratios for the specified angles, simplifying the fractions where possible.

🎯 Exam Tip: Always double-check which angle the ratio is being calculated for and correctly identify its opposite, adjacent, and hypotenuse sides within the relevant right-angled triangle.

Question 3. If \( 2 \cos \theta = \sqrt{3} \), then find all the trigonometric ratios of angle \( \theta \).
Answer:
Given \( 2 \cos \theta = \sqrt{3} \).
So, \( \cos \theta = \frac{\sqrt{3}}{2} \).
We know that \( \cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} \).
Let's consider a right-angled triangle ABC, where angle B is \( \theta \).
Thus, the adjacent side (BC) can be \( \sqrt{3} \) and the hypotenuse (AC) can be 2.
We use the Pythagorean theorem to find the opposite side (AB):
\( AB^2 = AC^2 - BC^2 \)
\( AB^2 = 2^2 - (\sqrt{3})^2 \)
\( AB^2 = 4 - 3 \)
\( AB^2 = 1 \)
\( AB = \sqrt{1} \)
\( AB = 1 \)

Now we have all three sides: Opposite = 1, Adjacent = \( \sqrt{3} \), Hypotenuse = 2.
We can find all the trigonometric ratios:
\( \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{1}{2} \)
\( \cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{\sqrt{3}}{2} \) (Given)
\( \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{1}{\sqrt{3}} \)
\( \csc \theta = \frac{\text{Hypotenuse}}{\text{Opposite side}} = \frac{2}{1} = 2 \)
\( \sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent side}} = \frac{2}{\sqrt{3}} \)
\( \cot \theta = \frac{\text{Adjacent side}}{\text{Opposite side}} = \frac{\sqrt{3}}{1} = \sqrt{3} \)
In simple words: We start with the given cosine value to know two sides of a right triangle. Then, we use Pythagoras' theorem to find the third side. Once all three sides are known, we can write down all six basic trigonometry ratios for angle theta.

🎯 Exam Tip: Remember special angle values like for 30°, 45°, and 60° (which \( \theta \) is in this case) to quickly verify your calculated ratios.

Question 4. If \( \cos A = \frac{3}{5} \), then find the value of \( \frac{\sin A-\cos A}{2 \tan A} \)
Answer:
Given \( \cos A = \frac{3}{5} \).
In a right-angled triangle ABC, with angle A, we have:
Adjacent side (AB) = 3
Hypotenuse (AC) = 5
Using the Pythagorean theorem to find the opposite side (BC):
\( BC^2 = AC^2 - AB^2 \)
\( BC^2 = 5^2 - 3^2 \)
\( BC^2 = 25 - 9 \)
\( BC^2 = 16 \)
\( BC = \sqrt{16} \)
\( BC = 4 \)

Now, we find \( \sin A \) and \( \tan A \):
\( \sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{4}{5} \)
\( \tan A = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AB} = \frac{4}{3} \)

Next, substitute these values into the given expression:
\( \frac{\sin A - \cos A}{2 \tan A} = \frac{\frac{4}{5} - \frac{3}{5}}{2 \times \frac{4}{3}} \)
\( = \frac{\frac{4-3}{5}}{\frac{8}{3}} \)
\( = \frac{\frac{1}{5}}{\frac{8}{3}} \)
To divide by a fraction, we multiply by its reciprocal:
\( = \frac{1}{5} \times \frac{3}{8} \)
\( = \frac{3}{40} \)
So, the value of the expression is \( \frac{3}{40} \).
In simple words: We use the given cosine value to find the missing side of a right triangle. Then, we calculate sine and tangent for the angle. Finally, we put these values into the expression and solve it like a fraction problem.

🎯 Exam Tip: Always calculate \( \sin A \), \( \cos A \), and \( \tan A \) first, then substitute them into the main expression to avoid errors in complex calculations.

Question 5. If \( \cos A = \frac{2x}{1+x^2} \) then find the values of \( \sin A \) and \( \tan A \) in terms of x.
Answer:
Given \( \cos A = \frac{2x}{1+x^2} \).
In a right-angled triangle ABC, with angle A:
Adjacent side (AB) = \( 2x \)
Hypotenuse (AC) = \( 1+x^2 \)
Using the Pythagorean theorem to find the opposite side (BC):
\( BC^2 = AC^2 - AB^2 \)
\( BC^2 = (1+x^2)^2 - (2x)^2 \)
\( BC^2 = (1 + 2x^2 + x^4) - 4x^2 \)
\( BC^2 = 1 - 2x^2 + x^4 \)
\( BC^2 = (x^2 - 1)^2 \) or \( (1 - x^2)^2 \)
So, \( BC = \sqrt{(x^2 - 1)^2} \)
\( BC = |x^2 - 1| \)
We can assume \( x^2 - 1 \) for simplicity, but it can also be \( 1 - x^2 \) if \( x < 1 \). Let's use \( x^2 - 1 \).

Now, we find \( \sin A \) and \( \tan A \):
\( \sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{x^2 - 1}{x^2 + 1} \)
\( \tan A = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AB} = \frac{x^2 - 1}{2x} \)
Alternatively, if we take \( BC = 1 - x^2 \):
\( \sin A = \frac{1 - x^2}{1 + x^2} \)
\( \tan A = \frac{1 - x^2}{2x} \)
Both forms are correct depending on the value of x relative to 1. The solution often presents one form. It's good to know that \( (x^2-1)^2 = (1-x^2)^2 \).
In simple words: We are given the cosine of angle A in terms of 'x'. We use this and Pythagoras' theorem to find the third side of the triangle, also in terms of 'x'. Once all three sides are known, we write down the sine and tangent ratios using these expressions.

🎯 Exam Tip: When dealing with expressions like \( (x^2-1)^2 \), remember that \( \sqrt{(x^2-1)^2} \) can result in either \( x^2-1 \) or \( 1-x^2 \). Choose the positive value or state both possibilities if context allows.

Question 6. If \( \sin \theta = \frac{a}{\sqrt{a^2+b^2}} \) then show that \( b \sin \theta = a \cos \theta \).
Answer:
Given \( \sin \theta = \frac{a}{\sqrt{a^2+b^2}} \).
In a right-angled triangle ABC, with angle B as \( \theta \):
Opposite side (AC) = \( a \)
Hypotenuse (AB) = \( \sqrt{a^2+b^2} \)
Using the Pythagorean theorem to find the adjacent side (BC):
\( BC^2 = AB^2 - AC^2 \)
\( BC^2 = (\sqrt{a^2+b^2})^2 - a^2 \)
\( BC^2 = (a^2+b^2) - a^2 \)
\( BC^2 = b^2 \)
\( BC = \sqrt{b^2} \)
\( BC = b \)

Now, we find \( \cos \theta \):
\( \cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{BC}{AB} = \frac{b}{\sqrt{a^2+b^2}} \)

Let's check the Left Hand Side (L.H.S) of the equation \( b \sin \theta = a \cos \theta \):
L.H.S \( = b \sin \theta \)
\( = b \times \frac{a}{\sqrt{a^2+b^2}} \)
\( = \frac{ab}{\sqrt{a^2+b^2}} \)

Now, let's check the Right Hand Side (R.H.S):
R.H.S \( = a \cos \theta \)
\( = a \times \frac{b}{\sqrt{a^2+b^2}} \)
\( = \frac{ab}{\sqrt{a^2+b^2}} \)

Since L.H.S = R.H.S, we have shown that \( b \sin \theta = a \cos \theta \). This identity is true for the given sine value.
In simple words: We use the given sine value and Pythagoras' theorem to find the missing side of the triangle. Then we calculate the cosine value. Finally, we put these sine and cosine values into both sides of the equation and show that they are equal.

🎯 Exam Tip: For "prove that" questions, always work with L.H.S and R.H.S separately and show they simplify to the same expression.

Question 7. If \( 3 \cot A = 2 \), then find the value of \( \frac{4\sin A-3\cos A}{2 \sin A+3 \cos A} \)
Answer:
Given \( 3 \cot A = 2 \).
So, \( \cot A = \frac{2}{3} \).
We know that \( \cot A = \frac{\text{Adjacent side}}{\text{Opposite side}} \).
In a right-angled triangle, with angle A:
Adjacent side (AB) = 2
Opposite side (BC) = 3
Using the Pythagorean theorem to find the hypotenuse (AC):
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = 2^2 + 3^2 \)
\( AC^2 = 4 + 9 \)
\( AC^2 = 13 \)
\( AC = \sqrt{13} \)

Now, we find \( \sin A \) and \( \cos A \):
\( \sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{3}{\sqrt{13}} \)
\( \cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{2}{\sqrt{13}} \)

Next, substitute these values into the given expression:
\( \frac{4\sin A-3\cos A}{2 \sin A+3 \cos A} = \frac{4(\frac{3}{\sqrt{13}}) - 3(\frac{2}{\sqrt{13}})}{2(\frac{3}{\sqrt{13}}) + 3(\frac{2}{\sqrt{13}})} \)
\( = \frac{\frac{12}{\sqrt{13}} - \frac{6}{\sqrt{13}}}{\frac{6}{\sqrt{13}} + \frac{6}{\sqrt{13}}} \)
\( = \frac{\frac{12-6}{\sqrt{13}}}{\frac{6+6}{\sqrt{13}}} \)
\( = \frac{\frac{6}{\sqrt{13}}}{\frac{12}{\sqrt{13}}} \)
\( = \frac{6}{\sqrt{13}} \times \frac{\sqrt{13}}{12} \)
\( = \frac{6}{12} \)
\( = \frac{1}{2} \)
In simple words: We first find the cotangent of A, then use it to draw a right triangle and find all its sides. From there, we calculate sine A and cosine A. Finally, we put these values into the given complex fraction and simplify it to get the answer.

🎯 Exam Tip: For expressions involving fractions with square roots in the denominator, you can often cancel out the common denominator \( (\sqrt{13} \) in this case) from numerator and denominator, simplifying the calculation.

Question 8. If \( \cos \theta : \sin \theta = 1 : 2 \), then find the value of \( \frac{8\cos \theta-2\sin \theta}{4 \cos \theta+2 \sin \theta} \)
Answer:
Given \( \cos \theta : \sin \theta = 1 : 2 \).
This means \( \frac{\cos \theta}{\sin \theta} = \frac{1}{2} \).
We know that \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), so \( \cot \theta = \frac{1}{2} \).
Alternatively, we can write \( \sin \theta = 2 \cos \theta \).

Now, let's substitute \( \sin \theta = 2 \cos \theta \) into the given expression:
\( \frac{8\cos \theta - 2\sin \theta}{4 \cos \theta + 2 \sin \theta} = \frac{8\cos \theta - 2(2 \cos \theta)}{4 \cos \theta + 2(2 \cos \theta)} \)
\( = \frac{8\cos \theta - 4 \cos \theta}{4 \cos \theta + 4 \cos \theta} \)
\( = \frac{4 \cos \theta}{8 \cos \theta} \)
Now, we can cancel \( \cos \theta \) from the numerator and denominator:
\( = \frac{4}{8} \)
\( = \frac{1}{2} \)

Alternatively, we can divide the numerator and denominator by \( \sin \theta \):
\( \frac{8\cos \theta - 2\sin \theta}{4 \cos \theta + 2 \sin \theta} = \frac{\frac{8\cos \theta}{\sin \theta} - \frac{2\sin \theta}{\sin \theta}}{\frac{4\cos \theta}{\sin \theta} + \frac{2\sin \theta}{\sin \theta}} \)
\( = \frac{8 \cot \theta - 2}{4 \cot \theta + 2} \)
Substitute \( \cot \theta = \frac{1}{2} \):
\( = \frac{8(\frac{1}{2}) - 2}{4(\frac{1}{2}) + 2} \)
\( = \frac{4 - 2}{2 + 2} \)
\( = \frac{2}{4} \)
\( = \frac{1}{2} \)
In simple words: We are given the ratio of cosine to sine. We can use this to express sine in terms of cosine (or vice versa). Then we replace sine in the main expression, simplify it, and cancel out common terms to get the final answer. Dividing by sine theta is another valid approach.

🎯 Exam Tip: When given a ratio of trigonometric functions, try to express one in terms of the other or convert the entire expression into terms of tangent or cotangent to simplify it easily.

Question 9. From the given figure, prove that \( \theta + \phi = 90^\circ \). Also prove that there are two other right angled triangles. Find \( \sin \alpha, \cos \beta \) and \( \tan \phi \).
Answer:
From the figure, we have a large triangle ABC, and a point D on AB, creating two smaller triangles ADC and BDC.
Given dimensions: AC = 15, BC = 20, AD = 9, DB = 16, CD = 12.

Let's first check if \( \triangle ABC \) is a right-angled triangle:
\( AB = AD + DB = 9 + 16 = 25 \)
Now, let's check Pythagoras theorem for \( \triangle ABC \):
\( AC^2 + BC^2 = 15^2 + 20^2 = 225 + 400 = 625 \)
\( AB^2 = 25^2 = 625 \)
Since \( AC^2 + BC^2 = AB^2 \), \( \triangle ABC \) is a right-angled triangle at C. Therefore, \( \angle C = 90^\circ \).
From the figure, \( \angle C = \theta + \phi \).
Since \( \angle C = 90^\circ \), it means \( \theta + \phi = 90^\circ \). (Proved)

Next, let's check for other right-angled triangles:
1. For \( \triangle ADC \):
\( AD^2 + CD^2 = 9^2 + 12^2 = 81 + 144 = 225 \)
\( AC^2 = 15^2 = 225 \)
Since \( AD^2 + CD^2 = AC^2 \), \( \triangle ADC \) is a right-angled triangle at D. Therefore, \( \angle ADC = 90^\circ \).

2. For \( \triangle BDC \):
\( BD^2 + CD^2 = 16^2 + 12^2 = 256 + 144 = 400 \)
\( BC^2 = 20^2 = 400 \)
Since \( BD^2 + CD^2 = BC^2 \), \( \triangle BDC \) is a right-angled triangle at D. Therefore, \( \angle BDC = 90^\circ \).
So, there are two other right-angled triangles: \( \triangle ADC \) and \( \triangle BDC \). (Proved)

Now, find \( \sin \alpha, \cos \beta \) and \( \tan \phi \):
In \( \triangle ADC \) (right-angled at D):
For angle \( \alpha \): Opposite side = CD = 12, Hypotenuse = AC = 15
\( \sin \alpha = \frac{CD}{AC} = \frac{12}{15} = \frac{4}{5} \)

In \( \triangle BDC \) (right-angled at D):
For angle \( \beta \): Adjacent side = BD = 16, Hypotenuse = BC = 20
\( \cos \beta = \frac{BD}{BC} = \frac{16}{20} = \frac{4}{5} \)

In \( \triangle ADC \) (right-angled at D):
For angle \( \phi \): Opposite side = AD = 9, Adjacent side = CD = 12
\( \tan \phi = \frac{AD}{CD} = \frac{9}{12} = \frac{3}{4} \)
In simple words: We first use the lengths of the sides to prove that the main triangle ABC is right-angled at C, which means \( \theta + \phi = 90^\circ \). Then we use Pythagoras' theorem for the smaller triangles to show that \( \triangle ADC \) and \( \triangle BDC \) are also right-angled at D. Finally, we use the side lengths to calculate \( \sin \alpha \), \( \cos \beta \), and \( \tan \phi \).

🎯 Exam Tip: Always verify if triangles are right-angled using the Pythagorean theorem before applying trigonometric ratios to them. This prevents errors in ratio selection.

Question 10. A boy standing at a point O finds his kite flying at a point P with distance OP = 25 m. It is at a height of 5 m from the ground. When the thread is extended by 10 m, it reaches a point Q. What will be the height QN of the kite from the ground? (use trigonometric ratios).
Answer:
Let the angle of elevation from point O be \( \theta \).
Initially, the kite is at point P. The distance from O to P (hypotenuse) is OP = 25 m. The height from the ground (opposite side) is PM = 5 m.
In the right-angled triangle \( \triangle OMP \):
\( \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{PM}{OP} = \frac{5}{25} = \frac{1}{5} \)

When the thread is extended by 10 m, the new distance from O to Q (hypotenuse) is OQ = OP + 10 = 25 + 10 = 35 m. The new height is QN (opposite side). The angle of elevation \( \theta \) remains the same.
In the right-angled triangle \( \triangle ONQ \):
\( \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{QN}{OQ} = \frac{QN}{35} \)

Since \( \sin \theta \) is the same in both cases:
\( \frac{QN}{35} = \frac{1}{5} \)
Now, we can solve for QN:
\( QN = \frac{1}{5} \times 35 \)
\( QN = 7 \)
So, the height of the kite from the ground at point Q is 7 m.
In simple words: We use the first position of the kite to find the sine of the angle at which the boy is looking up. Since the angle does not change, we use this same sine value with the new length of the kite string to find the kite's new height.

🎯 Exam Tip: In problems where the angle of elevation or depression remains constant, the trigonometric ratio (sine, cosine, or tangent) will also remain constant, making it easy to find unknown lengths.