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

 

Question 1.
(i) Prove that sin 60° = 2 sin 30° cos 30°.
(ii) Without using trigonometric tables find the value of 3 sin² 45° + 2 cos² 60°.
Answer:
(i) To prove: \( \sin 60^{\circ} = 2 \sin 30^{\circ} \cos 30^{\circ} \)
We know the standard trigonometric values:
\( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)
\( \sin 30^{\circ} = \frac{1}{2} \)
\( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)

Let's calculate the Left Hand Side (L.H.S.):
L.H.S. \( = \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)

Now, calculate the Right Hand Side (R.H.S.):
R.H.S. \( = 2 \sin 30^{\circ} \cos 30^{\circ} \)
\( = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} \)
\( = \frac{\sqrt{3}}{2} \)

Since L.H.S. \( = \frac{\sqrt{3}}{2} \) and R.H.S. \( = \frac{\sqrt{3}}{2} \), we can say:
\( \sin 60^{\circ} = 2 \sin 30^{\circ} \cos 30^{\circ} \)
This proves the statement. The trigonometric identities are consistent.

(ii) We need to find the value of \( 3 \sin^2 45^{\circ} + 2 \cos^2 60^{\circ} \).
We know the standard trigonometric values:
\( \sin 45^{\circ} = \frac{1}{\sqrt{2}} \)
\( \cos 60^{\circ} = \frac{1}{2} \)

Now, substitute these values into the expression:
\( 3 \sin^2 45^{\circ} + 2 \cos^2 60^{\circ} \)
\( = 3 \left( \frac{1}{\sqrt{2}} \right)^2 + 2 \left( \frac{1}{2} \right)^2 \)
\( = 3 \left( \frac{1}{2} \right) + 2 \left( \frac{1}{4} \right) \)
\( = \frac{3}{2} + \frac{2}{4} \)
\( = \frac{3}{2} + \frac{1}{2} \)
\( = \frac{3+1}{2} \)
\( = \frac{4}{2} \)
\( = 2 \)
The value of the expression is 2. This shows how knowing basic angle values helps in calculations.
In simple words: For the first part, we showed that the sine of 60 degrees is the same as two times the sine of 30 degrees multiplied by the cosine of 30 degrees, using known values. For the second part, we plugged in the values for sine 45 degrees and cosine 60 degrees into the given math problem and calculated the final answer to be 2.

🎯 Exam Tip: Memorizing the standard trigonometric ratios for common angles like 0°, 30°, 45°, 60°, and 90° is essential for solving such problems quickly and accurately without using tables.

 

Question 2. If \( 0 \le x \le 90^{\circ} \), state the numerical value of x for which \( \sin x^{\circ} = \cos x^{\circ} \).
Answer:
We are given the condition \( \sin x^{\circ} = \cos x^{\circ} \), where \( 0 \le x \le 90^{\circ} \).
To solve this, we can divide both sides by \( \cos x^{\circ} \) (assuming \( \cos x^{\circ} \ne 0 \)).
\( \frac{\sin x^{\circ}}{\cos x^{\circ}} = 1 \)

We know that \( \frac{\sin x^{\circ}}{\cos x^{\circ}} = \tan x^{\circ} \).
So, \( \tan x^{\circ} = 1 \)

We also know that the tangent of \( 45^{\circ} \) is 1.
\( \tan 45^{\circ} = 1 \)

Comparing \( \tan x^{\circ} = 1 \) with \( \tan 45^{\circ} = 1 \), we can find the value of x.
\( x = 45^{\circ} \)
This angle is the only one in the first quadrant where sine and cosine have the same value. This identity is a key concept in trigonometry.
In simple words: We want to find an angle where sine and cosine are equal. By dividing sine by cosine, we get tangent. Since tangent of 45 degrees is 1, the angle we are looking for is 45 degrees.

🎯 Exam Tip: Remember that \( \sin \theta = \cos \theta \) only when \( \theta = 45^{\circ} \) (or angles related to it by symmetry/periodicity) within the range \( 0^{\circ} \le \theta \le 90^{\circ} \). This is a common special angle relationship.

 

Question 3. Find the value of :
(i) sin² 60° + cos² 45°
(ii) 3 cos² 30° + tan² 60°
(iii) 4 sin² 60° + 3 tan² 30° – 8 sin 45° cos 45°
(iv) 2 sin² 30° – 3 cos² 45° + tan² 60°
(v) \( \frac{\sin 60^{\circ}}{\cos^2 45^{\circ}} - 3 \tan 30^{\circ} + 5 \cos 90^{\circ} \)
(vi) cos 90° + cos² 45° sin 30° tan 45°
(vii) cos² 45° + sin² 60° + sin² 30°
(viii) sin² 30° + cos² 60°
(ix) \( \frac{\sin^2 45^{\circ} + \cos^2 45^{\circ}}{\tan^2 60^{\circ}} \)
(x) \( \frac{5 \sin^2 30^{\circ} + \cos^2 45^{\circ} - 4 \tan^2 30^{\circ}}{2 \sin 30^{\circ} \cos 30^{\circ} + \tan 45^{\circ}} \)
(xi) \( 2\sqrt{2} \cos 45^{\circ} \cos 60^{\circ} + 2\sqrt{3} \sin 30^{\circ} \tan 60^{\circ} - \cos 0^{\circ} \)
(xii) \( \frac{4}{3} \tan^2 30^{\circ} + \sin^2 60^{\circ} - 3 \cos^2 60^{\circ} + \frac{3}{4} \tan^2 60^{\circ} - 2 \tan^2 45^{\circ} \)
(xiii) \( (\cos 0^{\circ} + \sin 45^{\circ} + \sin 30^{\circ}) \times (\sin 90^{\circ} - \cos 45^{\circ} + \cos 60^{\circ}) \)
Answer:
To solve these, we use the standard trigonometric values:
\( \sin 0^{\circ} = 0, \cos 0^{\circ} = 1, \tan 0^{\circ} = 0 \)
\( \sin 30^{\circ} = \frac{1}{2}, \cos 30^{\circ} = \frac{\sqrt{3}}{2}, \tan 30^{\circ} = \frac{1}{\sqrt{3}} \)
\( \sin 45^{\circ} = \frac{1}{\sqrt{2}}, \cos 45^{\circ} = \frac{1}{\sqrt{2}}, \tan 45^{\circ} = 1 \)
\( \sin 60^{\circ} = \frac{\sqrt{3}}{2}, \cos 60^{\circ} = \frac{1}{2}, \tan 60^{\circ} = \sqrt{3} \)
\( \sin 90^{\circ} = 1, \cos 90^{\circ} = 0, \tan 90^{\circ} = \text{undefined} \)

(i) \( \sin^2 60^{\circ} + \cos^2 45^{\circ} \)
\( = \left( \frac{\sqrt{3}}{2} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2 \)
\( = \frac{3}{4} + \frac{1}{2} \)
\( = \frac{3}{4} + \frac{2}{4} \)
\( = \frac{3+2}{4} \)
\( = \frac{5}{4} \)
\( = 1 \frac{1}{4} \)
This calculation involves squaring the values of sine and cosine for specific angles.

(ii) \( 3 \cos^2 30^{\circ} + \tan^2 60^{\circ} \)
\( = 3 \left( \frac{\sqrt{3}}{2} \right)^2 + (\sqrt{3})^2 \)
\( = 3 \times \frac{3}{4} + 3 \)
\( = \frac{9}{4} + 3 \)
\( = \frac{9}{4} + \frac{12}{4} \)
\( = \frac{9+12}{4} \)
\( = \frac{21}{4} \)
\( = 5 \frac{1}{4} \)
Here, we use the values of cosine and tangent for 30 and 60 degrees, respectively.

(iii) \( 4 \sin^2 60^{\circ} + 3 \tan^2 30^{\circ} - 8 \sin 45^{\circ} \cos 45^{\circ} \)
\( = 4 \left( \frac{\sqrt{3}}{2} \right)^2 + 3 \left( \frac{1}{\sqrt{3}} \right)^2 - 8 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \)
\( = 4 \times \frac{3}{4} + 3 \times \frac{1}{3} - 8 \times \frac{1}{2} \)
\( = 3 + 1 - 4 \)
\( = 4 - 4 \)
\( = 0 \)
This problem demonstrates how to combine squares of trigonometric ratios with products.

(iv) \( 2 \sin^2 30^{\circ} - 3 \cos^2 45^{\circ} + \tan^2 60^{\circ} \)
\( = 2 \left( \frac{1}{2} \right)^2 - 3 \left( \frac{1}{\sqrt{2}} \right)^2 + (\sqrt{3})^2 \)
\( = 2 \times \frac{1}{4} - 3 \times \frac{1}{2} + 3 \)
\( = \frac{1}{2} - \frac{3}{2} + 3 \)
\( = \frac{1 - 3 + 6}{2} \)
\( = \frac{4}{2} \)
\( = 2 \)
It's important to simplify fractions correctly in such multi-step problems.

(v) \( \frac{\sin 60^{\circ}}{\cos^2 45^{\circ}} - 3 \tan 30^{\circ} + 5 \cos 90^{\circ} \)
\( = \frac{\frac{\sqrt{3}}{2}}{\left( \frac{1}{\sqrt{2}} \right)^2} - 3 \times \frac{1}{\sqrt{3}} + 5 \times 0 \)
\( = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} - \frac{3}{\sqrt{3}} + 0 \)
\( = \frac{\sqrt{3}}{2} \times \frac{2}{1} - \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \)
\( = \sqrt{3} - \frac{3\sqrt{3}}{3} \)
\( = \sqrt{3} - \sqrt{3} \)
\( = 0 \)
Rationalizing the denominator of the fraction \( \frac{3}{\sqrt{3}} \) is a useful technique here.

(vi) \( \cos 90^{\circ} + \cos^2 45^{\circ} \sin 30^{\circ} \tan 45^{\circ} \)
\( = 0 + \left( \frac{1}{\sqrt{2}} \right)^2 \times \frac{1}{2} \times 1 \)
\( = 0 + \frac{1}{2} \times \frac{1}{2} \times 1 \)
\( = 0 + \frac{1}{4} \)
\( = \frac{1}{4} \)
Remember that \( \cos 90^{\circ} = 0 \), which simplifies the start of the expression significantly.

(vii) \( \cos^2 45^{\circ} + \sin^2 60^{\circ} + \sin^2 30^{\circ} \)
\( = \left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 + \left( \frac{1}{2} \right)^2 \)
\( = \frac{1}{2} + \frac{3}{4} + \frac{1}{4} \)
\( = \frac{2}{4} + \frac{3}{4} + \frac{1}{4} \)
\( = \frac{2+3+1}{4} \)
\( = \frac{6}{4} \)
\( = \frac{3}{2} \)
\( = 1 \frac{1}{2} \)
The identity \( \sin^2 \theta + \cos^2 \theta = 1 \) could have been used here for \( \sin^2 60^{\circ} + \cos^2 60^{\circ} \) if it were present, but here it's different angles.

(viii) \( \sin^2 30^{\circ} + \cos^2 60^{\circ} \)
\( = \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 \)
\( = \frac{1}{4} + \frac{1}{4} \)
\( = \frac{2}{4} \)
\( = \frac{1}{2} \)
This is a straightforward application of the values for 30 and 60 degrees.

(ix) \( \frac{\sin^2 45^{\circ} + \cos^2 45^{\circ}}{\tan^2 60^{\circ}} \)
\( = \frac{\left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2}{(\sqrt{3})^2} \)
\( = \frac{\frac{1}{2} + \frac{1}{2}}{3} \)
\( = \frac{1}{3} \)
The numerator uses the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), which makes it 1.

(x) \( \frac{5 \sin^2 30^{\circ} + \cos^2 45^{\circ} - 4 \tan^2 30^{\circ}}{2 \sin 30^{\circ} \cos 30^{\circ} + \tan 45^{\circ}} \)
Numerator: \( 5 \left( \frac{1}{2} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2 - 4 \left( \frac{1}{\sqrt{3}} \right)^2 \)
\( = 5 \times \frac{1}{4} + \frac{1}{2} - 4 \times \frac{1}{3} \)
\( = \frac{5}{4} + \frac{1}{2} - \frac{4}{3} \)
\( = \frac{15}{12} + \frac{6}{12} - \frac{16}{12} \)
\( = \frac{15+6-16}{12} \)
\( = \frac{5}{12} \)

Denominator: \( 2 \sin 30^{\circ} \cos 30^{\circ} + \tan 45^{\circ} \)
\( = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} + 1 \)
\( = \frac{\sqrt{3}}{2} + 1 \)
\( = \frac{\sqrt{3}+2}{2} \)

Now, divide the numerator by the denominator:
\( \frac{\frac{5}{12}}{\frac{\sqrt{3}+2}{2}} \)
\( = \frac{5}{12} \times \frac{2}{\sqrt{3}+2} \)
\( = \frac{5}{6(\sqrt{3}+2)} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{3}-2}{\sqrt{3}-2} \):
\( = \frac{5(\sqrt{3}-2)}{6((\sqrt{3})^2 - 2^2)} \)
\( = \frac{5(\sqrt{3}-2)}{6(3 - 4)} \)
\( = \frac{5(\sqrt{3}-2)}{6(-1)} \)
\( = \frac{5(2-\sqrt{3})}{6} \)
This problem requires careful calculation of both the numerator and the denominator separately.

(xi) \( 2\sqrt{2} \cos 45^{\circ} \cos 60^{\circ} + 2\sqrt{3} \sin 30^{\circ} \tan 60^{\circ} - \cos 0^{\circ} \)
\( = 2\sqrt{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{2} + 2\sqrt{3} \times \frac{1}{2} \times \sqrt{3} - 1 \)
\( = (2 \times \frac{1}{2}) + (2 \times \frac{1}{2} \times 3) - 1 \)
\( = 1 + 3 - 1 \)
\( = 3 \)
This expression simplifies nicely after substituting the trigonometric values.

(xii) \( \frac{4}{3} \tan^2 30^{\circ} + \sin^2 60^{\circ} - 3 \cos^2 60^{\circ} + \frac{3}{4} \tan^2 60^{\circ} - 2 \tan^2 45^{\circ} \)
\( = \frac{4}{3} \left( \frac{1}{\sqrt{3}} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 - 3 \left( \frac{1}{2} \right)^2 + \frac{3}{4} (\sqrt{3})^2 - 2 (1)^2 \)
\( = \frac{4}{3} \times \frac{1}{3} + \frac{3}{4} - 3 \times \frac{1}{4} + \frac{3}{4} \times 3 - 2 \times 1 \)
\( = \frac{4}{9} + \frac{3}{4} - \frac{3}{4} + \frac{9}{4} - 2 \)
\( = \frac{4}{9} + \frac{9}{4} - 2 \)
\( = \frac{16}{36} + \frac{81}{36} - \frac{72}{36} \)
\( = \frac{16+81-72}{36} \)
\( = \frac{25}{36} \)
Make sure to find a common denominator when adding and subtracting fractions.

(xiii) \( (\cos 0^{\circ} + \sin 45^{\circ} + \sin 30^{\circ}) \times (\sin 90^{\circ} - \cos 45^{\circ} + \cos 60^{\circ}) \)
First parenthesis: \( \cos 0^{\circ} + \sin 45^{\circ} + \sin 30^{\circ} = 1 + \frac{1}{\sqrt{2}} + \frac{1}{2} = \frac{2+\sqrt{2}+1}{2} = \frac{3+\sqrt{2}}{2} \)
Second parenthesis: \( \sin 90^{\circ} - \cos 45^{\circ} + \cos 60^{\circ} = 1 - \frac{1}{\sqrt{2}} + \frac{1}{2} = \frac{2-\sqrt{2}+1}{2} = \frac{3-\sqrt{2}}{2} \)
Now multiply the results:
\( \left( \frac{3+\sqrt{2}}{2} \right) \times \left( \frac{3-\sqrt{2}}{2} \right) \)
\( = \frac{(3+\sqrt{2})(3-\sqrt{2})}{2 \times 2} \)
Using the identity \( (a+b)(a-b) = a^2 - b^2 \):
\( = \frac{3^2 - (\sqrt{2})^2}{4} \)
\( = \frac{9 - 2}{4} \)
\( = \frac{7}{4} \)
\( = 1 \frac{3}{4} \)
This problem conveniently uses the difference of squares identity for multiplication.
In simple words: We substituted the known values of sine, cosine, and tangent for special angles (0, 30, 45, 60, 90 degrees) into each expression. Then, we carefully followed the order of operations (like squaring and multiplying) and simplified the fractions to find the final numerical answer for each part. It's like solving small puzzles using a set of rules for numbers in trigonometry.

🎯 Exam Tip: When evaluating complex trigonometric expressions, break them down into smaller, manageable parts. Calculate each term separately, then combine them. Pay close attention to signs, powers, and order of operations to avoid errors. Also, use common denominators carefully for fractions.

 

Question 4. ABC is an isosceles right triangle. Assuming AB = BC = x, find the value of each of the following trigonometric ratios.
(i) sin 45°
(ii) cos 45°
(iii) tan 45°
Show your complete working neatly.
Answer:
In triangle ABC, it is an isosceles right-angled triangle. This means two sides are equal, and one angle is \( 90^{\circ} \). The problem statement says assume AB = BC = x, but the solution proceeds with AB = AC = x and the diagram shows Angle A = \( 90^{\circ} \), Angle B = \( 45^{\circ} \), Angle C = \( 45^{\circ} \). We will follow the geometry from the image and the solution's first line: In \( \triangle ABC \), \( \angle A = 90^{\circ} \), and AB = AC = x.
Since \( \angle A = 90^{\circ} \) and AB = AC, then \( \angle B = \angle C = 45^{\circ} \).
Draw AD perpendicular to BC. In an isosceles right triangle, this altitude bisects the angle A and also the hypotenuse BC.

By Pythagoras theorem in \( \triangle ABC \):
\( BC^2 = AB^2 + AC^2 \)
\( BC^2 = x^2 + x^2 = 2x^2 \)
\( BC = \sqrt{2x^2} = x\sqrt{2} \)

Since AD bisects BC, \( BD = DC = \frac{BC}{2} = \frac{x\sqrt{2}}{2} \).
In \( \triangle ABD \), which is a right-angled triangle at D:
\( AD^2 + BD^2 = AB^2 \)
\( AD^2 + \left( \frac{x\sqrt{2}}{2} \right)^2 = x^2 \)
\( AD^2 + \frac{2x^2}{4} = x^2 \)
\( AD^2 + \frac{x^2}{2} = x^2 \)
\( AD^2 = x^2 - \frac{x^2}{2} = \frac{x^2}{2} \)
\( AD = \sqrt{\frac{x^2}{2}} = \frac{x}{\sqrt{2}} = \frac{x\sqrt{2}}{2} \)
So, \( AD = BD = CD = \frac{x\sqrt{2}}{2} \). This makes \( \triangle ABD \) and \( \triangle ACD \) isosceles right triangles too.

Let's find the trigonometric ratios for \( 45^{\circ} \) using \( \triangle ABD \) or \( \triangle ABC \). We'll use \( \angle B = 45^{\circ} \).
(i) \( \sin 45^{\circ} \)
In \( \triangle ABD \), for \( \angle B \), the opposite side is AD, and the hypotenuse is AB.
\( \sin B = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AD}{AB} \)
\( \sin 45^{\circ} = \frac{\frac{x\sqrt{2}}{2}}{x} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \)

(ii) \( \cos 45^{\circ} \)
In \( \triangle ABD \), for \( \angle B \), the adjacent side is BD, and the hypotenuse is AB.
\( \cos B = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BD}{AB} \)
\( \cos 45^{\circ} = \frac{\frac{x\sqrt{2}}{2}}{x} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \)

(iii) \( \tan 45^{\circ} \)
In \( \triangle ABD \), for \( \angle B \), the opposite side is AD, and the adjacent side is BD.
\( \tan B = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AD}{BD} \)
\( \tan 45^{\circ} = \frac{\frac{x\sqrt{2}}{2}}{\frac{x\sqrt{2}}{2}} = 1 \)
These calculations confirm the standard trigonometric values for 45 degrees, which are fundamental.
In simple words: We used a right-angled triangle where two sides are equal (making the other two angles 45 degrees each). By finding the lengths of all sides using Pythagoras theorem, we could then calculate sine, cosine, and tangent for the 45-degree angle. These calculations confirmed that sine 45 and cosine 45 are both \( \frac{1}{\sqrt{2}} \) and tangent 45 is 1.

🎯 Exam Tip: When dealing with geometric problems involving trigonometric ratios, always draw and label a clear diagram. Identify the right-angled triangles and apply SOH CAH TOA (Sine=Opposite/Hypotenuse, Cosine=Adjacent/Hypotenuse, Tangent=Opposite/Adjacent) correctly to the relevant angles.

 

Question 5. Without using tables, find the value of
\( \frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \times \tan 60^{\circ}} \)
Answer:
We need to find the value of the given expression without using trigonometric tables.
We will use the standard trigonometric values:
\( \sin 30^{\circ} = \frac{1}{2} \)
\( \sin 90^{\circ} = 1 \)
\( \cos 0^{\circ} = 1 \)
\( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \)
\( \tan 60^{\circ} = \sqrt{3} \)

Substitute these values into the expression:
\( \frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \times \tan 60^{\circ}} \)
\( = \frac{\frac{1}{2} - 1 + 2 \times 1}{\frac{1}{\sqrt{3}} \times \sqrt{3}} \)
\( = \frac{\frac{1}{2} - 1 + 2}{1} \)
\( = \frac{\frac{1}{2} + 1}{1} \)
\( = \frac{1}{2} + 1 \)
\( = \frac{1}{2} + \frac{2}{2} \)
\( = \frac{3}{2} \)
\( = 1.5 \)
The value of the expression is 1.5. This calculation emphasizes the importance of recalling exact trigonometric values for special angles.
In simple words: We replaced all the sine, cosine, and tangent parts with their exact number values. Then, we did the addition, subtraction, and multiplication steps to get the final answer, which is 1.5.

🎯 Exam Tip: Simplify the numerator and the denominator separately before performing the final division. This reduces the chances of calculation errors. Also, be careful with the order of operations, especially multiplication and division.

 

Question 6. Without using tables verify that :
(i) \( \sin 60^{\circ} = \frac{2 \tan 30^{\circ}}{1+\tan^2 30^{\circ}} \)
(ii) \( \cos 60^{\circ} = \frac{1-\tan^2 30^{\circ}}{1+\tan^2 30^{\circ}} \)
(iii) \( \cos 60^{\circ} = \cos^2 30^{\circ} - \sin^2 30^{\circ} \)
(iv) \( \cos 60^{\circ} = 1 - 2 \sin^2 30^{\circ} = 2 \cos^2 30^{\circ} - 1 \)
Answer:
We need to verify the given trigonometric identities by substituting the standard values of the angles without using tables.
Standard trigonometric values:
\( \sin 30^{\circ} = \frac{1}{2}, \cos 30^{\circ} = \frac{\sqrt{3}}{2}, \tan 30^{\circ} = \frac{1}{\sqrt{3}} \)
\( \sin 60^{\circ} = \frac{\sqrt{3}}{2}, \cos 60^{\circ} = \frac{1}{2} \)

(i) Verify \( \sin 60^{\circ} = \frac{2 \tan 30^{\circ}}{1+\tan^2 30^{\circ}} \)
L.H.S. \( = \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)

R.H.S. \( = \frac{2 \tan 30^{\circ}}{1+\tan^2 30^{\circ}} \)
\( = \frac{2 \times \frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^2} \)
\( = \frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} \)
\( = \frac{2}{\sqrt{3}} \times \frac{3}{4} \)
\( = \frac{6}{4\sqrt{3}} \)
\( = \frac{3}{2\sqrt{3}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{3}}{\sqrt{3}} \):
\( = \frac{3\sqrt{3}}{2 \times 3} \)
\( = \frac{\sqrt{3}}{2} \)
Since L.H.S. = R.H.S. \( = \frac{\sqrt{3}}{2} \), the identity is verified. This formula is a useful double-angle identity for sine.

(ii) Verify \( \cos 60^{\circ} = \frac{1-\tan^2 30^{\circ}}{1+\tan^2 30^{\circ}} \)
L.H.S. \( = \cos 60^{\circ} = \frac{1}{2} \)

R.H.S. \( = \frac{1-\tan^2 30^{\circ}}{1+\tan^2 30^{\circ}} \)
\( = \frac{1-\left(\frac{1}{\sqrt{3}}\right)^2}{1+\left(\frac{1}{\sqrt{3}}\right)^2} \)
\( = \frac{1-\frac{1}{3}}{1+\frac{1}{3}} \)
\( = \frac{\frac{3-1}{3}}{\frac{3+1}{3}} \)
\( = \frac{\frac{2}{3}}{\frac{4}{3}} \)
\( = \frac{2}{3} \times \frac{3}{4} \)
\( = \frac{2}{4} \)
\( = \frac{1}{2} \)
Since L.H.S. = R.H.S. \( = \frac{1}{2} \), the identity is verified. This is another form of the double-angle identity for cosine.

(iii) Verify \( \cos 60^{\circ} = \cos^2 30^{\circ} - \sin^2 30^{\circ} \)
L.H.S. \( = \cos 60^{\circ} = \frac{1}{2} \)

R.H.S. \( = \cos^2 30^{\circ} - \sin^2 30^{\circ} \)
\( = \left( \frac{\sqrt{3}}{2} \right)^2 - \left( \frac{1}{2} \right)^2 \)
\( = \frac{3}{4} - \frac{1}{4} \)
\( = \frac{3-1}{4} \)
\( = \frac{2}{4} \)
\( = \frac{1}{2} \)
Since L.H.S. = R.H.S. \( = \frac{1}{2} \), the identity is verified. This is also a common double-angle identity for cosine.

(iv) Verify \( \cos 60^{\circ} = 1 - 2 \sin^2 30^{\circ} \) and \( \cos 60^{\circ} = 2 \cos^2 30^{\circ} - 1 \)
L.H.S. \( = \cos 60^{\circ} = \frac{1}{2} \)

First part R.H.S.: \( 1 - 2 \sin^2 30^{\circ} \)
\( = 1 - 2 \left( \frac{1}{2} \right)^2 \)
\( = 1 - 2 \times \frac{1}{4} \)
\( = 1 - \frac{1}{2} \)
\( = \frac{1}{2} \)
This matches the L.H.S., so the first part is verified.

Second part R.H.S.: \( 2 \cos^2 30^{\circ} - 1 \)
\( = 2 \left( \frac{\sqrt{3}}{2} \right)^2 - 1 \)
\( = 2 \times \frac{3}{4} - 1 \)
\( = \frac{3}{2} - 1 \)
\( = \frac{1}{2} \)
This also matches the L.H.S., so the second part is verified. These are further variations of the double-angle identity for cosine, useful in different contexts.
In simple words: We checked four different math rules to see if they are true. For each rule, we put in the exact number values for angles like 30 degrees and 60 degrees. After doing the calculations on both sides of each rule, we found that the numbers matched, meaning all four rules are correct. These rules help us understand how angles relate to each other in trigonometry.

🎯 Exam Tip: When verifying trigonometric identities, always work on one side (usually the more complex one) and transform it to match the other side. Do not cross the equality sign unless you are confident it's an identity. Clearly show each step of substitution and simplification.

 

Question 7. Prove that:
(i) cos² 30° + sin² 30° + tan² 45° = 2.
(ii) 4 (sin⁴ 30° + cos⁴ 60°) – 3 (cos² 45° – sin² 90°) = 2.
Answer:
To prove these statements, we will substitute the standard trigonometric values:
\( \sin 30^{\circ} = \frac{1}{2}, \cos 30^{\circ} = \frac{\sqrt{3}}{2}, \tan 45^{\circ} = 1 \)
\( \sin 90^{\circ} = 1, \cos 60^{\circ} = \frac{1}{2}, \cos 45^{\circ} = \frac{1}{\sqrt{2}} \)

(i) Prove \( \cos^2 30^{\circ} + \sin^2 30^{\circ} + \tan^2 45^{\circ} = 2 \)
L.H.S. \( = \cos^2 30^{\circ} + \sin^2 30^{\circ} + \tan^2 45^{\circ} \)
We know that \( \cos^2 \theta + \sin^2 \theta = 1 \). So, \( \cos^2 30^{\circ} + \sin^2 30^{\circ} = 1 \).
\( = 1 + (\tan 45^{\circ})^2 \)
\( = 1 + (1)^2 \)
\( = 1 + 1 \)
\( = 2 \)
Since L.H.S. = 2 and R.H.S. = 2, the identity is proven. This shows the power of using fundamental trigonometric identities.

(ii) Prove \( 4 (\sin^4 30^{\circ} + \cos^4 60^{\circ}) - 3 (\cos^2 45^{\circ} - \sin^2 90^{\circ}) = 2 \)
L.H.S. \( = 4 (\sin^4 30^{\circ} + \cos^4 60^{\circ}) - 3 (\cos^2 45^{\circ} - \sin^2 90^{\circ}) \)
\( = 4 \left( \left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^4 \right) - 3 \left( \left(\frac{1}{\sqrt{2}}\right)^2 - (1)^2 \right) \)
\( = 4 \left( \frac{1}{16} + \frac{1}{16} \right) - 3 \left( \frac{1}{2} - 1 \right) \)
\( = 4 \left( \frac{2}{16} \right) - 3 \left( -\frac{1}{2} \right) \)
\( = 4 \times \frac{1}{8} - 3 \times \left(-\frac{1}{2}\right) \)
\( = \frac{4}{8} + \frac{3}{2} \)
\( = \frac{1}{2} + \frac{3}{2} \)
\( = \frac{1+3}{2} \)
\( = \frac{4}{2} \)
\( = 2 \)
Since L.H.S. = 2 and R.H.S. = 2, the identity is proven. This problem involves higher powers of trigonometric ratios and careful arithmetic.
In simple words: For both parts, we replaced the sine, cosine, and tangent terms with their specific number values. Then, we did all the math steps like squaring, raising to the power of four, adding, and subtracting. In both cases, the final answer came out to be 2, which matched what we needed to prove. This confirms the given equations are correct.

🎯 Exam Tip: When dealing with powers higher than 2 (like \( \sin^4 \theta \)), calculate the base trigonometric value first, then raise it to the given power. Remember to handle negative signs carefully, especially when subtracting. Always check if a known identity (like \( \sin^2 \theta + \cos^2 \theta = 1 \)) can simplify any part of the expression.

 

Question 8. A pole 15 m long rests against a vertical wall at an angle of 60° with the ground. Calculate
(i) how high up the wall will the pole reach?
(ii) how far is the foot of the pole from the wall?
Answer:
Let AB represent the vertical wall, BC represent the ground, and AC represent the pole.
The pole (AC) is 15 m long.
The pole makes an angle of \( 60^{\circ} \) with the ground, so \( \angle C = 60^{\circ} \).
Since the wall is vertical to the ground, \( \angle B = 90^{\circ} \).

Let 'x' be the height the pole reaches up the wall (AB).
Let 'y' be the distance of the foot of the pole from the wall (BC).

(i) To find the height 'x' (AB):
In right-angled \( \triangle ABC \):
We know the hypotenuse AC = 15 m and \( \angle C = 60^{\circ} \). We want to find the opposite side AB.
We use the sine ratio: \( \sin C = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} \)
\( \sin 60^{\circ} = \frac{x}{15} \)
We know \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \).
\( \frac{\sqrt{3}}{2} = \frac{x}{15} \)

\( \implies x = \frac{15\sqrt{3}}{2} \)
Given \( \sqrt{3} \approx 1.732 \).
\( x = \frac{15 \times 1.732}{2} \)
\( x = \frac{25.980}{2} \)
\( x = 12.99 \text{ m} \)
The height the pole reaches up the wall is 12.99 m. This is a common application of trigonometry in real-world scenarios.

(ii) To find the distance 'y' (BC):
In right-angled \( \triangle ABC \):
We know the hypotenuse AC = 15 m and \( \angle C = 60^{\circ} \). We want to find the adjacent side BC.
We use the cosine ratio: \( \cos C = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} \)
\( \cos 60^{\circ} = \frac{y}{15} \)
We know \( \cos 60^{\circ} = \frac{1}{2} \).
\( \frac{1}{2} = \frac{y}{15} \)

\( \implies y = \frac{15}{2} \)
\( y = 7.5 \text{ m} \)
The distance of the foot of the pole from the wall is 7.5 m. This illustrates how trigonometry helps in determining unknown distances.
In simple words: We have a pole leaning against a wall, making a triangle. To find how high the pole reaches on the wall, we used the sine function with the angle the pole makes with the ground. To find how far the pole's base is from the wall, we used the cosine function with the same angle. We plugged in the known values for the angle and pole length to get the height as 12.99 meters and the distance as 7.5 meters.

🎯 Exam Tip: Always draw a diagram for word problems involving heights and distances. Clearly label the known and unknown values and the angles. Choose the appropriate trigonometric ratio (sine, cosine, or tangent) based on the sides you know and the side you need to find relative to the given angle.

 

Question 9. In the given triangle, calculate \( \theta \) if B = 90°, AB = 20 cm and AC = 40 cm.
Answer:
We are given a right-angled triangle ABC, with \( \angle B = 90^{\circ} \).
The length of side AB (opposite to \( \angle C \)) is 20 cm.
The length of the hypotenuse AC is 40 cm.
We need to find the value of angle \( \theta \) which is \( \angle C \).

In \( \triangle ABC \):
\( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} \)
\( \sin \theta = \frac{20}{40} \)
\( \sin \theta = \frac{1}{2} \)

We know that \( \sin 30^{\circ} = \frac{1}{2} \).
Therefore, by comparing the values:
\( \theta = 30^{\circ} \)
The angle \( \theta \) is 30 degrees. This is a direct application of the sine ratio to find an angle.
In simple words: We have a right-angled triangle where we know the length of the side opposite angle theta and the hypotenuse. We used the sine function, which is opposite side divided by hypotenuse, and found that \( \sin \theta \) equals 1/2. Since \( \sin 30^{\circ} \) is also 1/2, the angle \( \theta \) must be 30 degrees.

🎯 Exam Tip: When given two sides of a right-angled triangle and asked to find an angle, determine which trigonometric ratio (sine, cosine, or tangent) relates the given sides to the unknown angle. Then, use the inverse trigonometric function (e.g., \( \sin^{-1} \)) or recall standard values to find the angle.

 

Question 10. Without using tables solve each of the following triangle ABC, right-angled at C; given
(i) A = 30° and c = 40;
(ii) B = 60° and c = 15;
(iii) A = 45° and a = 7.
Answer:
For a triangle ABC right-angled at C, we have \( \angle C = 90^{\circ} \).
Sides: 'a' is opposite \( \angle A \), 'b' is opposite \( \angle B \), and 'c' is opposite \( \angle C \) (hypotenuse).

(i) Given: \( \angle A = 30^{\circ} \) and hypotenuse \( c = 40 \).
Since \( \angle C = 90^{\circ} \) and \( \angle A = 30^{\circ} \):
\( \angle B = 180^{\circ} - (\angle A + \angle C) = 180^{\circ} - (30^{\circ} + 90^{\circ}) = 180^{\circ} - 120^{\circ} = 60^{\circ} \).

To find side 'a' (opposite \( \angle A \)):
\( \sin A = \frac{a}{c} \)
\( \sin 30^{\circ} = \frac{a}{40} \)
\( \frac{1}{2} = \frac{a}{40} \)

\( \implies a = \frac{40}{2} = 20 \).

To find side 'b' (adjacent to \( \angle A \)):
\( \cos A = \frac{b}{c} \)
\( \cos 30^{\circ} = \frac{b}{40} \)
\( \frac{\sqrt{3}}{2} = \frac{b}{40} \)

\( \implies b = \frac{40\sqrt{3}}{2} = 20\sqrt{3} \)
Using \( \sqrt{3} \approx 1.732 \), \( b = 20 \times 1.732 = 34.64 \).
So, \( \angle B = 60^{\circ} \), \( a = 20 \), \( b = 34.64 \). Solving a triangle means finding all unknown angles and sides.

(ii) Given: \( \angle B = 60^{\circ} \) and hypotenuse \( c = 15 \).
Since \( \angle C = 90^{\circ} \) and \( \angle B = 60^{\circ} \):
\( \angle A = 180^{\circ} - (\angle B + \angle C) = 180^{\circ} - (60^{\circ} + 90^{\circ}) = 180^{\circ} - 150^{\circ} = 30^{\circ} \).

To find side 'b' (opposite \( \angle B \)):
\( \sin B = \frac{b}{c} \)
\( \sin 60^{\circ} = \frac{b}{15} \)
\( \frac{\sqrt{3}}{2} = \frac{b}{15} \)

\( \implies b = \frac{15\sqrt{3}}{2} \)
Using \( \sqrt{3} \approx 1.732 \), \( b = \frac{15 \times 1.732}{2} = \frac{25.98}{2} = 12.99 \).

To find side 'a' (adjacent to \( \angle B \)):
\( \cos B = \frac{a}{c} \)
\( \cos 60^{\circ} = \frac{a}{15} \)
\( \frac{1}{2} = \frac{a}{15} \)

\( \implies a = \frac{15}{2} = 7.5 \).
So, \( \angle A = 30^{\circ} \), \( a = 7.5 \), \( b = 12.99 \). This part also uses complementary angles within a right triangle.

(iii) Given: \( \angle A = 45^{\circ} \) and side \( a = 7 \).
Since \( \angle C = 90^{\circ} \) and \( \angle A = 45^{\circ} \):
\( \angle B = 180^{\circ} - (\angle A + \angle C) = 180^{\circ} - (45^{\circ} + 90^{\circ}) = 180^{\circ} - 135^{\circ} = 45^{\circ} \).
Since \( \angle A = \angle B = 45^{\circ} \), this is an isosceles right triangle, meaning \( a = b \).
So, \( b = 7 \).

To find hypotenuse 'c':
\( \sin A = \frac{a}{c} \)
\( \sin 45^{\circ} = \frac{7}{c} \)
\( \frac{1}{\sqrt{2}} = \frac{7}{c} \)

\( \implies c = 7\sqrt{2} \)
Using \( \sqrt{2} \approx 1.414 \), \( c = 7 \times 1.414 = 9.898 \approx 9.9 \).
So, \( \angle B = 45^{\circ} \), \( b = 7 \), \( c = 9.9 \). It's helpful to recognize when a triangle is isosceles to quickly find side lengths.
In simple words: For each part of the question, we used the given angle and side in a right-angled triangle. We first found the third angle using the fact that angles in a triangle add up to 180 degrees. Then, we used sine and cosine rules to find the lengths of the other two sides. For example, if we knew one angle and the hypotenuse, we used sine to find the opposite side and cosine to find the adjacent side.

🎯 Exam Tip: When solving a triangle, always start by finding the missing angle if two are known. Then, use the sine or cosine ratios (SOH CAH TOA) to find the unknown sides. Remember that the sum of angles in a triangle is \( 180^{\circ} \).

 

Question 11. If 4 sin² θ – 1 = 0 and angle θ is less than 90°; find the value of θ and hence the value of cos² θ + tan² θ.
Answer:
Given \( 4 \sin^2 \theta - 1 = 0 \).
Add 1 to both sides: \( 4 \sin^2 \theta = 1 \).
Divide by 4: \( \sin^2 \theta = \frac{1}{4} \).
Take the square root of both sides: \( \sin \theta = \sqrt{\frac{1}{4}} = \frac{1}{2} \).
Since \( \theta \) is an acute angle and \( \sin \theta = \frac{1}{2} \), we know that \( \theta = 30^\circ \). The angle 30 degrees is often found using a 30-60-90 right triangle.
Now, we need to find the value of \( \cos^2 \theta + \tan^2 \theta \) for \( \theta = 30^\circ \).
\( \cos^2 30^\circ + \tan^2 30^\circ \)
Substitute the known values: \( (\frac{\sqrt{3}}{2})^2 + (\frac{1}{\sqrt{3}})^2 \).
\( = \frac{3}{4} + \frac{1}{3} \).
To add these fractions, find a common denominator, which is 12.
\( = \frac{3 \times 3}{4 \times 3} + \frac{1 \times 4}{3 \times 4} \).
\( = \frac{9}{12} + \frac{4}{12} \).
\( = \frac{9+4}{12} = \frac{13}{12} \).
The final answer can also be written as a mixed number, \( 1\frac{1}{12} \).
In simple words: First, we solve the equation to find that the angle \( \theta \) is 30 degrees. Then, we put this angle into the expression for cosine and tangent, and calculate the final number.

🎯 Exam Tip: Remember the exact trigonometric values for standard angles like 30°, 45°, and 60°. Always solve for the angle first, then substitute it into the expression to be evaluated.

 

Question 12. If θ is an acute angle and sin 0 = cos 0, find the value of 2 tan² 0 + sin² 0 – 1.
Answer:
Given that \( \sin \theta = \cos \theta \).
To find \( \theta \), divide both sides by \( \cos \theta \) (since \( \cos \theta \neq 0 \) for an acute angle where \( \sin \theta = \cos \theta \)).
\( \frac{\sin \theta}{\cos \theta} = 1 \).
We know that \( \frac{\sin \theta}{\cos \theta} = \tan \theta \).
So, \( \tan \theta = 1 \).
For an acute angle, \( \tan \theta = 1 \) implies \( \theta = 45^\circ \). This means the angle is 45 degrees, where sine and cosine values are equal.
Now, we need to find the value of \( 2 \tan^2 \theta + \sin^2 \theta - 1 \) for \( \theta = 45^\circ \).
\( 2 \tan^2 45^\circ + \sin^2 45^\circ - 1 \).
Substitute the known values: \( 2(1)^2 + (\frac{1}{\sqrt{2}})^2 - 1 \).
\( = 2(1) + \frac{1}{2} - 1 \).
\( = 2 + \frac{1}{2} - 1 \).
\( = 1 + \frac{1}{2} \).
\( = \frac{3}{2} \) or \( 1.5 \).
In simple words: When \( \sin \theta \) equals \( \cos \theta \) for an acute angle, \( \theta \) must be 45 degrees. Then, we put 45 degrees into the expression and calculate the final number.

🎯 Exam Tip: The identity \( \sin \theta = \cos \theta \) always means \( \tan \theta = 1 \) (for valid \( \theta \)), which quickly gives \( \theta = 45^\circ \). Knowing this can save time in calculations.

 

Question 13. A kite, flying at a height of 75 metres from the level ground, is attached to a string inclined at 60° to the horizontal, find the length of the string to the nearest metre. (Take √3 = 1.73)
Answer:
Let the height of the kite from the ground be \( h = 75 \) m.
Let the length of the string be \( x \) m.
The angle of elevation (angle the string makes with the horizontal) is \( 60^\circ \).
We can form a right-angled triangle where the height is the opposite side to the angle, and the string length is the hypotenuse.
Using the sine trigonometric ratio: \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \).
\( \sin 60^\circ = \frac{h}{x} \).
Substitute the values: \( \frac{\sqrt{3}}{2} = \frac{75}{x} \).
Now, solve for \( x \): \( x\sqrt{3} = 75 \times 2 \).
\( x\sqrt{3} = 150 \).
\( x = \frac{150}{\sqrt{3}} \).
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} \):
\( x = \frac{150 \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{150 \sqrt{3}}{3} \).
\( x = 50 \sqrt{3} \).
Given \( \sqrt{3} = 1.73 \).
\( x = 50 \times 1.73 \).
\( x = 86.5 \) m.
Rounding to the nearest metre, the length of the string is \( 87 \) m. A kite string must be strong enough to withstand wind forces, which is why they are often made of durable materials.
In simple words: We draw a right triangle where the kite's height is one side and the string is the longest side. Using the sine function with the given angle and height, we calculate the length of the string.

🎯 Exam Tip: Always draw a diagram for height and distance problems to visualize the right-angled triangle and correctly identify the opposite, adjacent, and hypotenuse sides relative to the given angle.

 

Question 14. If 2 sin 0 – 1 = 0, find : (i) the value of 0 in degrees where O is an acute angle; (ii) cos20 + tan20.
Answer:
(i) Find the value of \( \theta \):
Given \( 2 \sin \theta - 1 = 0 \).
Add 1 to both sides: \( 2 \sin \theta = 1 \).
Divide by 2: \( \sin \theta = \frac{1}{2} \).
Since \( \theta \) is an acute angle, \( \sin \theta = \frac{1}{2} \) means \( \theta = 30^\circ \). This is a fundamental trigonometric value often memorized.
(ii) Find the value of \( \cos^2 \theta + \tan^2 \theta \):
Substitute \( \theta = 30^\circ \) into the expression:
\( \cos^2 30^\circ + \tan^2 30^\circ \).
Using standard trigonometric values: \( \cos 30^\circ = \frac{\sqrt{3}}{2} \) and \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
\( = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 \).
\( = \frac{3}{4} + \frac{1}{3} \).
To add these fractions, find the least common multiple of 4 and 3, which is 12.
\( = \frac{3 \times 3}{4 \times 3} + \frac{1 \times 4}{3 \times 4} \).
\( = \frac{9}{12} + \frac{4}{12} \).
\( = \frac{9+4}{12} = \frac{13}{12} \).
The value is \( \frac{13}{12} \) or \( 1\frac{1}{12} \).
In simple words: First, we solve the given equation to find that the angle \( \theta \) is 30 degrees. Then, we use this angle to calculate the value of the second expression by plugging in the known sine and tangent values.

🎯 Exam Tip: Ensure you use the correct angle (found in part i) for the second calculation. Double-check your squaring of fractions and finding common denominators when adding them.

 

Question 15. The altitude AD of a △ABC, in which ∠A is obtuse, is 10 cm. If BD = 10 cm and CD = 10√3 cm, determine ∠A.
Answer:
Let AD be the altitude of \( \triangle ABC \) to BC, with AD = 10 cm.
Given BD = 10 cm and CD = \( 10\sqrt{3} \) cm.
We have two right-angled triangles: \( \triangle ABD \) and \( \triangle ACD \).
In \( \triangle ABD \):
\( \tan(\angle BAD) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BD}{AD} \).
\( \tan(\angle BAD) = \frac{10}{10} = 1 \).
Therefore, \( \angle BAD = 45^\circ \).
In \( \triangle ACD \):
\( \tan(\angle CAD) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{CD}{AD} \).
\( \tan(\angle CAD) = \frac{10\sqrt{3}}{10} = \sqrt{3} \).
Therefore, \( \angle CAD = 60^\circ \).
Since \( \angle A \) is obtuse and AD is an altitude within the triangle, \( \angle A \) is the sum of \( \angle BAD \) and \( \angle CAD \). An obtuse angle is greater than 90 degrees.
\( \angle A = \angle BAD + \angle CAD \).
\( \angle A = 45^\circ + 60^\circ \).
\( \angle A = 105^\circ \).
In simple words: We split the obtuse angle A into two smaller angles using the altitude. By using the tangent function in each smaller right triangle, we find the values of these two angles and then add them together to get the total angle A.

🎯 Exam Tip: When dealing with an obtuse angle in a triangle with an internal altitude, remember that the obtuse angle is the sum of the two angles formed at the vertex by the altitude. Sketching the triangle is essential.

 

Question 16. In a right triangle ABC, right angled at C, if ∠B = 60° and AB = 15 units, find the remaining angles and sides.
Answer:
Given: Right triangle ABC, with \( \angle C = 90^\circ \), \( \angle B = 60^\circ \), and hypotenuse AB = 15 units.
First, find the remaining angle \( \angle A \):
The sum of angles in a triangle is \( 180^\circ \).
\( \angle A + \angle B + \angle C = 180^\circ \).
\( \angle A + 60^\circ + 90^\circ = 180^\circ \).
\( \angle A + 150^\circ = 180^\circ \).
\( \angle A = 180^\circ - 150^\circ = 30^\circ \).
Next, find the lengths of the remaining sides, AC and BC.
To find AC (opposite to \( \angle B \)):
Use \( \sin B = \frac{AC}{AB} \).
\( \sin 60^\circ = \frac{AC}{15} \).
\( \frac{\sqrt{3}}{2} = \frac{AC}{15} \).
\( AC = 15 \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{2} \) units.
To find BC (adjacent to \( \angle B \)):
Use \( \cos B = \frac{BC}{AB} \).
\( \cos 60^\circ = \frac{BC}{15} \).
\( \frac{1}{2} = \frac{BC}{15} \).
\( BC = 15 \times \frac{1}{2} = 7.5 \) units.
So, the remaining angle is \( \angle A = 30^\circ \), and the remaining sides are \( AC = \frac{15\sqrt{3}}{2} \) units and \( BC = 7.5 \) units. The use of exact trigonometric values provides precise answers.
In simple words: We find the third angle of the triangle by subtracting the known angles from 180 degrees. Then, we use the sine and cosine rules with the given hypotenuse and angles to find the lengths of the other two sides.

🎯 Exam Tip: Always start by finding the third angle using the angle sum property of a triangle. Then, use SOH CAH TOA (Sine, Cosine, Tangent) to find the unknown sides, picking the ratio that uses a known side and angle to find an unknown side.

 

Question 17. In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.
Answer:
In a rectangle ABCD, all angles are \( 90^\circ \). Consider the right-angled triangle ABC, where \( \angle B = 90^\circ \).
Given AB = 20 cm and \( \angle BAC = 60^\circ \).
To calculate side BC (opposite to \( \angle BAC \)):
We use the tangent ratio: \( \tan(\angle BAC) = \frac{BC}{AB} \).
\( \tan 60^\circ = \frac{BC}{20} \).
We know \( \tan 60^\circ = \sqrt{3} \).
\( \sqrt{3} = \frac{BC}{20} \).
\( BC = 20\sqrt{3} \) cm.
To calculate diagonal AC (hypotenuse of \( \triangle ABC \)):
We use the cosine ratio: \( \cos(\angle BAC) = \frac{AB}{AC} \).
\( \cos 60^\circ = \frac{20}{AC} \).
We know \( \cos 60^\circ = \frac{1}{2} \).
\( \frac{1}{2} = \frac{20}{AC} \).
\( AC = 20 \times 2 = 40 \) cm.
In a rectangle, the diagonals are equal in length. So, diagonal BD = diagonal AC.
\( BD = 40 \) cm.
Therefore, side BC = \( 20\sqrt{3} \) cm, diagonal AC = 40 cm, and diagonal BD = 40 cm. Understanding the properties of rectangles simplifies these calculations.
In simple words: We use the given side and angle in the right-angled corner of the rectangle to find the other side using tangent and the diagonal using cosine. Since diagonals in a rectangle are always equal, we get the length of the second diagonal immediately.

🎯 Exam Tip: Remember that a rectangle has four right angles and its diagonals are equal. This allows you to apply right-triangle trigonometry (SOH CAH TOA) to find unknown sides and diagonals.

 

Question 18. A ladder is placed against a wall such that it just reaches the top of the wall. The foot of the ladder is 1.5 m away from the wall and the ladder is inclined at an angle of 60° with the ground. Find the height of the wall. (Given √3 = 1.73)
Answer:
Let the height of the wall be \( h \) (opposite side).
The distance of the foot of the ladder from the wall is 1.5 m (adjacent side).
The angle of inclination of the ladder with the ground is \( 60^\circ \).
We can form a right-angled triangle where the wall, the ground, and the ladder are the sides.
To find the height of the wall, we use the tangent trigonometric ratio: \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \).
\( \tan 60^\circ = \frac{h}{1.5} \).
We know that \( \tan 60^\circ = \sqrt{3} \).
So, \( \sqrt{3} = \frac{h}{1.5} \).
Multiply both sides by 1.5: \( h = 1.5 \sqrt{3} \).
Given \( \sqrt{3} = 1.73 \).
\( h = 1.5 \times 1.73 \).
\( h = 2.595 \) m.
Therefore, the height of the wall is 2.595 m. Real-world problems like this demonstrate the practical application of trigonometry.
In simple words: We create a right triangle with the wall, ground, and ladder. Using the angle the ladder makes with the ground and the distance from the wall, we use the tangent function to find the height of the wall.

🎯 Exam Tip: For problems involving heights and distances, always draw a neat diagram to represent the situation. This helps in correctly identifying the right-angled triangle and choosing the appropriate trigonometric ratio (sine, cosine, or tangent).

 

Question 19. An electric pole is 10 m high. A steel wire tied to the top of the pole is affixed at a point on the ground to keep the pole upright. If the steel wire makes an angle 45° with the horizontal through the foot of the pole, find the length of the steel wire. (Given √2 = 1.41)
Answer:
Let the height of the electric pole be \( h = 10 \) m.
Let the length of the steel wire be \( x \) m (hypotenuse).
The angle the wire makes with the horizontal (angle of elevation) is \( 45^\circ \).
We can form a right-angled triangle where the pole is the opposite side to the angle, and the wire is the hypotenuse.
To find the length of the steel wire, we use the sine trigonometric ratio: \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \).
\( \sin 45^\circ = \frac{h}{x} \).
Substitute the values: \( \frac{1}{\sqrt{2}} = \frac{10}{x} \).
Now, solve for \( x \): \( x = 10 \sqrt{2} \).
Given \( \sqrt{2} = 1.41 \).
\( x = 10 \times 1.41 \).
\( x = 14.1 \) m.
Therefore, the length of the steel wire is 14.1 m. Using a steel wire at this angle provides stability to the pole.
In simple words: We draw a right triangle with the pole as one side and the wire as the longest side. Using the sine function with the pole's height and the angle, we can find the length of the wire.

🎯 Exam Tip: Remember that for a 45° angle, \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \). This specific angle simplifies calculations significantly in many trigonometry problems.

 

Question 20. A vertically straight tree, 15 m high, is broken by the wind in such a way that its top just touches the ground and makes an angle of 60° with the ground. At what height from the ground did the tree break ? (Use √3 = 1.73)
Answer:
Let the total height of the tree be \( H = 15 \) m.
Let the height from the ground where the tree broke be \( x \) m. This is the standing part of the tree.
The broken part of the tree is \( (15 - x) \) m. This broken part forms the hypotenuse of the right-angled triangle.
When the top touches the ground, it forms an angle of \( 60^\circ \) with the ground.
In the right-angled triangle formed:
The opposite side to the \( 60^\circ \) angle is the standing height of the tree, \( x \).
The hypotenuse is the broken part of the tree, \( (15 - x) \).
Using the sine trigonometric ratio: \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \).
\( \sin 60^\circ = \frac{x}{15-x} \).
Substitute the value of \( \sin 60^\circ \): \( \frac{\sqrt{3}}{2} = \frac{x}{15-x} \).
Cross-multiply: \( \sqrt{3}(15-x) = 2x \).
\( 15\sqrt{3} - \sqrt{3}x = 2x \).
Move \( \sqrt{3}x \) to the right side: \( 15\sqrt{3} = 2x + \sqrt{3}x \).
Factor out \( x \): \( 15\sqrt{3} = x(2+\sqrt{3}) \).
Solve for \( x \): \( x = \frac{15\sqrt{3}}{2+\sqrt{3}} \).
To rationalize the denominator, multiply the numerator and denominator by the conjugate \( (2-\sqrt{3}) \):
\( x = \frac{15\sqrt{3}(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} \).
Using the identity \( (a+b)(a-b) = a^2-b^2 \) in the denominator:
\( x = \frac{15\sqrt{3} \times 2 - 15\sqrt{3} \times \sqrt{3}}{2^2 - (\sqrt{3})^2} \).
\( x = \frac{30\sqrt{3} - 15 \times 3}{4 - 3} \).
\( x = \frac{30\sqrt{3} - 45}{1} \).
\( x = 30\sqrt{3} - 45 \).
Given \( \sqrt{3} = 1.73 \).
\( x = 30 \times 1.73 - 45 \).
\( x = 51.9 - 45 \).
\( x = 6.9 \) m.
The tree broke at a height of 6.9 m from the ground. This type of problem is a common application of right-angle trigonometry.
In simple words: The tree breaks into two parts; the top part falls and becomes the slanted side of a triangle, while the bottom part remains standing and is the height we need to find. We set up an equation using the sine function and solve for the unknown height.

🎯 Exam Tip: For "broken tree" problems, draw a clear diagram and remember that the sum of the standing part and the broken part equals the original height of the tree. The broken part acts as the hypotenuse in the right-angled triangle.

 

Question 21. The string of a kite is 150 m long and it makes an angle of 60° with the horizontal. Find the height of the kite from the ground. (Take √3 = 1.73).
Answer:
Let the length of the kite string be \( L = 150 \) m. This is the hypotenuse of the right triangle.
Let the height of the kite from the ground be \( h \) m. This is the side opposite to the angle of elevation.
The angle the string makes with the horizontal (angle of elevation) is \( 60^\circ \).
We can form a right-angled triangle where the kite's height is the opposite side and the string length is the hypotenuse.
To find the height of the kite, we use the sine trigonometric ratio: \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \).
\( \sin 60^\circ = \frac{h}{L} \).
Substitute the values: \( \frac{\sqrt{3}}{2} = \frac{h}{150} \).
Now, solve for \( h \): \( 2h = 150\sqrt{3} \).
\( h = \frac{150\sqrt{3}}{2} \).
\( h = 75\sqrt{3} \).
Given \( \sqrt{3} = 1.73 \).
\( h = 75 \times 1.73 \).
\( h = 129.75 \) m.
Therefore, the height of the kite from the ground is 129.75 m. This calculation shows how string length and angle determine flight height.
In simple words: We imagine a right triangle where the string is the slanted side and the kite's height is the straight up-and-down side. Using the sine function with the given angle and string length, we calculate how high the kite is.

🎯 Exam Tip: Always make a sketch for word problems involving heights and distances. Clearly label the knowns and unknowns, then choose the trigonometric ratio (sine, cosine, or tangent) that relates them correctly.