Class 10 Mathematics - Ch 8: Application of Trigonometry
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📘 Comprehensive Syllabus & Examination Guide
Class 10 Mathematics - Ch 8: Application of Trigonometry
Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
🎯 Question Types & Curriculum Breakdown
Total Question Pool
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116 Questions
Combined Active Syllabus
Short Questions
72%
84 Questions
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Multiple Choice (MCQs)
10%
12 MCQs
Available
Fill in the Blanks
9%
10 Questions
Available
True / False
7%
8 Questions
Available
Match Columns
2%
2 Questions
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📊 Question Pool Structure
116 Solved Questions (MCQs, Short & Long Questions, Blanks, True/False).
⚡ Recommended Pacing
1 to 3 minutes per question depending on question type (MCQ, Short, Long).
⚖️ Scoring & Negative Marking
1 to 5 marks per question aligned with official board examination rubrics.
💡 Strategic Preparation & Exam Hall Guidelines
To maximize your score on Class 10 Mathematics - Ch 8: Application of Trigonometry, candidates are advised to follow a structured three-pass approach. In the First Pass, solve all direct recall and formula-based questions within 30 seconds each to secure foundational marks. In the Second Pass, tackle multi-step analytical and quantitative reasoning problems. In the Third Pass, review marked questions and verify calculations.
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📝 Pre-Rendered Solved Sample Questions & Detailed Solutions
Review the solved problems below to understand question phrasing, answer choices, and step-by-step solution logic prior to starting the full interactive practice drill:
Sample Question 1
Exercise 8.1 - Reference Angles in Quadrant II
MEDIUM • Short Question
Find the reference angle of $110^\circ$.
✓ Correct Answer: 70°
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Identify the quadrant of the angle:</strong></p>
<p>The given angle $\theta = 110^\circ$ lies in the <strong>second quadrant</strong> ($90^\circ < 110^\circ < 180^\circ$).</p>
<p><strong>Step 2: Apply the reference angle formula for Quadrant II:</strong></p>
<p>$$\theta_{\text{ref}} = 180^\circ - \theta$$</p>
<p>$$\theta_{\text{ref}} = 180^\circ - 110^\circ = 70^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\theta_{\text{ref}} = 70^\circ}$$</p>
<p>The given angle $\theta = 110^\circ$ lies in the <strong>second quadrant</strong> ($90^\circ < 110^\circ < 180^\circ$).</p>
<p><strong>Step 2: Apply the reference angle formula for Quadrant II:</strong></p>
<p>$$\theta_{\text{ref}} = 180^\circ - \theta$$</p>
<p>$$\theta_{\text{ref}} = 180^\circ - 110^\circ = 70^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\theta_{\text{ref}} = 70^\circ}$$</p>
Sample Question 2
Exercise 8.1 - Reference Angles in Quadrant II
MEDIUM • Short Question
Find the reference angle of $138^\circ$.
✓ Correct Answer: 42°
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Identify the quadrant of the angle:</strong></p>
<p>The angle $\theta = 138^\circ$ lies in the <strong>second quadrant</strong> ($90^\circ < 138^\circ < 180^\circ$).</p>
<p><strong>Step 2: Apply the reference angle formula for Quadrant II:</strong></p>
<p>$$\theta_{\text{ref}} = 180^\circ - \theta$$</p>
<p>$$\theta_{\text{ref}} = 180^\circ - 138^\circ = 42^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\theta_{\text{ref}} = 42^\circ}$$</p>
<p>The angle $\theta = 138^\circ$ lies in the <strong>second quadrant</strong> ($90^\circ < 138^\circ < 180^\circ$).</p>
<p><strong>Step 2: Apply the reference angle formula for Quadrant II:</strong></p>
<p>$$\theta_{\text{ref}} = 180^\circ - \theta$$</p>
<p>$$\theta_{\text{ref}} = 180^\circ - 138^\circ = 42^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\theta_{\text{ref}} = 42^\circ}$$</p>
Sample Question 3
Exercise 8.1 - Reference Angles in Quadrant II
MEDIUM • Short Question
Find the reference angle of $125^\circ$.
✓ Correct Answer: 55°
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Identify the quadrant of the angle:</strong></p>
<p>The angle $\theta = 125^\circ$ lies in the <strong>second quadrant</strong> ($90^\circ < 125^\circ < 180^\circ$).</p>
<p><strong>Step 2: Apply the reference angle formula for Quadrant II:</strong></p>
<p>$$\theta_{\text{ref}} = 180^\circ - \theta$$</p>
<p>$$\theta_{\text{ref}} = 180^\circ - 125^\circ = 55^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\theta_{\text{ref}} = 55^\circ}$$</p>
<p>The angle $\theta = 125^\circ$ lies in the <strong>second quadrant</strong> ($90^\circ < 125^\circ < 180^\circ$).</p>
<p><strong>Step 2: Apply the reference angle formula for Quadrant II:</strong></p>
<p>$$\theta_{\text{ref}} = 180^\circ - \theta$$</p>
<p>$$\theta_{\text{ref}} = 180^\circ - 125^\circ = 55^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\theta_{\text{ref}} = 55^\circ}$$</p>
Sample Question 4
Exercise 8.1 - Reference Angles in Quadrant II
MEDIUM • Short Question
Find the reference angle of $142^\circ$.
✓ Correct Answer: 38°
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Identify the quadrant of the angle:</strong></p>
<p>The angle $\theta = 142^\circ$ lies in the <strong>second quadrant</strong> ($90^\circ < 142^\circ < 180^\circ$).</p>
<p><strong>Step 2: Apply the reference angle formula for Quadrant II:</strong></p>
<p>$$\theta_{\text{ref}} = 180^\circ - \theta$$</p>
<p>$$\theta_{\text{ref}} = 180^\circ - 142^\circ = 38^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\theta_{\text{ref}} = 38^\circ}$$</p>
<p>The angle $\theta = 142^\circ$ lies in the <strong>second quadrant</strong> ($90^\circ < 142^\circ < 180^\circ$).</p>
<p><strong>Step 2: Apply the reference angle formula for Quadrant II:</strong></p>
<p>$$\theta_{\text{ref}} = 180^\circ - \theta$$</p>
<p>$$\theta_{\text{ref}} = 180^\circ - 142^\circ = 38^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\theta_{\text{ref}} = 38^\circ}$$</p>
Sample Question 5
Exercise 8.1 - Equivalent Angle for Sine in Quadrant II
MEDIUM • Short Question
If $\theta = 36^\circ$, find the equivalent angle $\alpha$ in the second quadrant for which $\sin \alpha = \sin \theta$.
✓ Correct Answer: 144°
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Understand the trigonometric relation for sine in Quadrant II:</strong></p>
<p>In the unit circle, the sine function is positive in both the first and second quadrants. The relationship between an angle $\theta$ and its supplementary angle $\alpha$ in the second quadrant is:</p>
<p>$$\sin(180^\circ - \theta) = \sin \theta$$</p>
<p><strong>Step 2: Calculate the angle $\alpha$:</strong></p>
<p>$$\alpha = 180^\circ - \theta = 180^\circ - 36^\circ = 144^\circ$$</p>
<p><strong>Verification:</strong></p>
<p>$$\sin 144^\circ = \sin(180^\circ - 36^\circ) = \sin 36^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\alpha = 144^\circ}$$</p>
<p>In the unit circle, the sine function is positive in both the first and second quadrants. The relationship between an angle $\theta$ and its supplementary angle $\alpha$ in the second quadrant is:</p>
<p>$$\sin(180^\circ - \theta) = \sin \theta$$</p>
<p><strong>Step 2: Calculate the angle $\alpha$:</strong></p>
<p>$$\alpha = 180^\circ - \theta = 180^\circ - 36^\circ = 144^\circ$$</p>
<p><strong>Verification:</strong></p>
<p>$$\sin 144^\circ = \sin(180^\circ - 36^\circ) = \sin 36^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\alpha = 144^\circ}$$</p>
Sample Question 6
Exercise 8.1 - Equivalent Angle for Cosine in Quadrant II
MEDIUM • Short Question
If $\theta = 87^\circ$, find the equivalent angle $\beta$ in the second quadrant for which $\cos \beta = -\cos \theta$.
✓ Correct Answer: 93°
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Understand the trigonometric relation for cosine in Quadrant II:</strong></p>
<p>In the second quadrant, the cosine function is negative. For any acute angle $\theta$, the cosine of the supplementary angle $\beta = 180^\circ - \theta$ satisfies:</p>
<p>$$\cos(180^\circ - \theta) = -\cos \theta$$</p>
<p><strong>Step 2: Calculate the angle $\beta$:</strong></p>
<p>$$\beta = 180^\circ - \theta = 180^\circ - 87^\circ = 93^\circ$$</p>
<p><strong>Verification:</strong></p>
<p>$$\cos 93^\circ = \cos(180^\circ - 87^\circ) = -\cos 87^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\beta = 93^\circ}$$</p>
<p>In the second quadrant, the cosine function is negative. For any acute angle $\theta$, the cosine of the supplementary angle $\beta = 180^\circ - \theta$ satisfies:</p>
<p>$$\cos(180^\circ - \theta) = -\cos \theta$$</p>
<p><strong>Step 2: Calculate the angle $\beta$:</strong></p>
<p>$$\beta = 180^\circ - \theta = 180^\circ - 87^\circ = 93^\circ$$</p>
<p><strong>Verification:</strong></p>
<p>$$\cos 93^\circ = \cos(180^\circ - 87^\circ) = -\cos 87^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\beta = 93^\circ}$$</p>
Sample Question 7
Exercise 8.1 - Exact Values of Trigonometric Ratios of 120°
MEDIUM • Short Question
Using a reference angle, find the exact values of $\sin 120^\circ$, $\cos 120^\circ$, and $\tan 120^\circ$.
✓ Correct Answer: sin 120° = √3/2, cos 120° = -1/2, tan 120° = -√3
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Find the reference angle for $120^\circ$:</strong></p>
<p>The angle $120^\circ$ lies in Quadrant II. Its reference angle is:</p>
<p>$$\theta_{\text{ref}} = 180^\circ - 120^\circ = 60^\circ$$</p>
<p><strong>Step 2: Determine exact values for the reference angle $60^\circ$:</strong></p>
<p>$$\sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2}, \quad \tan 60^\circ = \sqrt{3}$$</p>
<p><strong>Step 3: Apply Quadrant II signs (Sine positive, Cosine & Tangent negative):</strong></p>
<p>$$\sin 120^\circ = \sin(180^\circ - 60^\circ) = +\sin 60^\circ = \frac{\sqrt{3}}{2}$$</p>
<p>$$\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}$$</p>
<p>$$\tan 120^\circ = \tan(180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\sin 120^\circ = \frac{\sqrt{3}}{2}, \quad \cos 120^\circ = -\frac{1}{2}, \quad \tan 120^\circ = -\sqrt{3}}$$</p>
<p>The angle $120^\circ$ lies in Quadrant II. Its reference angle is:</p>
<p>$$\theta_{\text{ref}} = 180^\circ - 120^\circ = 60^\circ$$</p>
<p><strong>Step 2: Determine exact values for the reference angle $60^\circ$:</strong></p>
<p>$$\sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2}, \quad \tan 60^\circ = \sqrt{3}$$</p>
<p><strong>Step 3: Apply Quadrant II signs (Sine positive, Cosine & Tangent negative):</strong></p>
<p>$$\sin 120^\circ = \sin(180^\circ - 60^\circ) = +\sin 60^\circ = \frac{\sqrt{3}}{2}$$</p>
<p>$$\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}$$</p>
<p>$$\tan 120^\circ = \tan(180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\sin 120^\circ = \frac{\sqrt{3}}{2}, \quad \cos 120^\circ = -\frac{1}{2}, \quad \tan 120^\circ = -\sqrt{3}}$$</p>
Sample Question 8
Exercise 8.1 - Exact Values for 3π/4 Radian Measure
MEDIUM • Short Question
Using the reference angle, find the exact values of $\cos\left(\frac{3\pi}{4}\right)$, $\sin\left(\frac{3\pi}{4}\right)$, and $\tan\left(\frac{3\pi}{4}\right)$.
✓ Correct Answer: cos(3π/4) = -1/√2, sin(3π/4) = 1/√2, tan(3π/4) = -1
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Convert/Identify the reference angle in radians:</strong></p>
<p>The angle $\frac{3\pi}{4} = 135^\circ$ lies in Quadrant II ($\frac{\pi}{2} < \frac{3\pi}{4} < \pi$).</p>
<p>$$\theta_{\text{ref}} = \pi - \frac{3\pi}{4} = \frac{\pi}{4} \quad (45^\circ)$$</p>
<p><strong>Step 2: Standard values for $\frac{\pi}{4}$:</strong></p>
<p>$$\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \tan\left(\frac{\pi}{4}\right) = 1$$</p>
<p><strong>Step 3: Evaluate ratios with Quadrant II signs:</strong></p>
<p>$$\sin\left(\frac{3\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$</p>
<p>$$\cos\left(\frac{3\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$$</p>
<p>$$\tan\left(\frac{3\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right) = -1$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}, \quad \sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \tan\left(\frac{3\pi}{4}\right) = -1}$$</p>
<p>The angle $\frac{3\pi}{4} = 135^\circ$ lies in Quadrant II ($\frac{\pi}{2} < \frac{3\pi}{4} < \pi$).</p>
<p>$$\theta_{\text{ref}} = \pi - \frac{3\pi}{4} = \frac{\pi}{4} \quad (45^\circ)$$</p>
<p><strong>Step 2: Standard values for $\frac{\pi}{4}$:</strong></p>
<p>$$\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \tan\left(\frac{\pi}{4}\right) = 1$$</p>
<p><strong>Step 3: Evaluate ratios with Quadrant II signs:</strong></p>
<p>$$\sin\left(\frac{3\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$</p>
<p>$$\cos\left(\frac{3\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$$</p>
<p>$$\tan\left(\frac{3\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right) = -1$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}, \quad \sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \tan\left(\frac{3\pi}{4}\right) = -1}$$</p>
Sample Question 9
Exercise 8.1 - Trigonometric Ratios of Supplementary Angles
MEDIUM • Short Question
If $\cos \theta = 0.559$, find the value of $\cos(180^\circ - \theta)$.
✓ Correct Answer: -0.559
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Use the reduction identity for cosine:</strong></p>
<p>$$\cos(180^\circ - \theta) = -\cos \theta$$</p>
<p><strong>Step 2: Substitute $\cos \theta = 0.559$:</strong></p>
<p>$$\cos(180^\circ - \theta) = -0.559$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{-0.559}$$</p>
<p>$$\cos(180^\circ - \theta) = -\cos \theta$$</p>
<p><strong>Step 2: Substitute $\cos \theta = 0.559$:</strong></p>
<p>$$\cos(180^\circ - \theta) = -0.559$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{-0.559}$$</p>
Sample Question 10
Exercise 8.1 - Trigonometric Ratios of Supplementary Angles
MEDIUM • Short Question
If $\cos \theta = 0.559$, find the value of $\sin(180^\circ - \theta)$.
✓ Correct Answer: 0.83
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Calculate $\sin \theta$ using the Pythagorean identity:</strong></p>
<p>$$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (0.559)^2} = \sqrt{1 - 0.312481} = \sqrt{0.687519} \approx 0.82916 \approx 0.83$$</p>
<p><strong>Step 2: Use the reduction identity for sine:</strong></p>
<p>$$\sin(180^\circ - \theta) = \sin \theta \approx 0.83$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{0.83}$$</p>
<p>$$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (0.559)^2} = \sqrt{1 - 0.312481} = \sqrt{0.687519} \approx 0.82916 \approx 0.83$$</p>
<p><strong>Step 2: Use the reduction identity for sine:</strong></p>
<p>$$\sin(180^\circ - \theta) = \sin \theta \approx 0.83$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{0.83}$$</p>
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