Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 8 Set 8.2 Trigonometry Solutions

Question 1. In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक तालिका दी गई है जिसमें त्रिकोणमितीय अनुपात sin θ, cos θ और tan θ की पंक्तियाँ हैं और i से ix तक स्तंभ हैं। प्रत्येक स्तंभ में एक अनुपात दिया गया है और शेष दो अनुपात ज्ञात कर तालिका को पूरा करना है। यह प्रश्न त्रिकोणमितीय अनुपातों के बीच संबंधों का उपयोग करके गुम मानों को भरने पर केंद्रित है।

Answer: Solution: i. cos \( \theta \) = \( \frac{35}{37} \) ...(i) [Given] In right angled \( \triangle \)ABC,
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC, जिसमें कोण C थीटा (θ) है, कोण B समकोण है, AC भुजा की लंबाई 37k है और BC भुजा की लंबाई 35k है। ∠C = \( \theta \). cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} \) cos \( \theta \) = \( \frac{BC}{AC} \) ...(ii)
\( \implies \frac{BC}{AC} = \frac{35}{37} \) ...[From (i) and (ii)] Let the common multiple be k. .: BC = 35k and AC = 37k Now, AC\(^2\) = AB\(^2\) + BC\(^2\) ...[Pythagoras theorem] .: (37k)\(^2\) = AB\(^2\) + (35k)\(^2\) 1369k\(^2\) = AB\(^2\) + 1225k\(^2\) AB\(^2\) = 1369k\(^2\) - 1225k\(^2\) = 144k\(^2\) AB = \( \sqrt{144k^2} \) ... [Taking square root of both sides] = 12k .: sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12k}{37k} = \frac{12}{37} \) tan \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Adjacent side of } \theta} = \frac{AB}{BC} = \frac{12k}{35k} = \frac{12}{35} \) ii. sin \( \theta \) = \( \frac{11}{61} \) ......(i) [Given] In right angled \( \triangle \)ABC, ∠C = \( \theta \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC, जिसमें कोण C थीटा (θ) है, कोण B समकोण है, AC भुजा की लंबाई 61k है और AB भुजा की लंबाई 11k है। sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} \) sin \( \theta \) = \( \frac{AB}{AC} \) ...(ii)
\( \implies \frac{AB}{AC} = \frac{11}{61} \) ...[From (i) and (ii)] Let the common multiple be k. AB = 11k and AC = 61k Now, AC\(^2\) = AB\(^2\) + BC\(^2\) ...[Pythagoras theorem] .: (61k)\(^2\) = (11k)\(^2\) + BC\(^2\) .: 3721k\(^2\) = 121k\(^2\) + BC\(^2\) .: BC\(^2\) = 3721k\(^2\) - 121k\(^2\) = 3600k\(^2\) BC = \( \sqrt{3600k^2} \) ...[Taking square root of both sides] = 60k .: cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{60k}{61k} = \frac{60}{61} \) tan \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Adjacent side of } \theta} = \frac{AB}{BC} = \frac{11k}{60k} = \frac{11}{60} \) iii. tan \( \theta \) = 1 = \( \frac{1}{1} \) ..(i) [Given] In right angled \( \triangle \)ABC, ∠C = \( \theta \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC, जिसमें कोण C थीटा (θ) है, कोण B समकोण है, AB भुजा की लंबाई 1k है और BC भुजा की लंबाई 1k है। tan \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Adjacent side of } \theta} \) tan \( \theta \) = \( \frac{AB}{BC} \) ...(ii)
\( \implies \frac{AB}{BC} = \frac{1}{1} \) ...[From (i) and (ii)] Let the common multiple be k. .: AB = 1k and BC = 1k Now, AC\(^2\) = AB\(^2\) + BC\(^2\) ...[Pythagoras theorem] = k\(^2\) + k\(^2\) = 2k\(^2\) .: AC = \( \sqrt{2k^2} \) ...[Taking square root of both sides] = \( \sqrt{2} \)k sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{1k}{\sqrt{2}k} = \frac{1}{\sqrt{2}} \) cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{1k}{\sqrt{2}k} = \frac{1}{\sqrt{2}} \) iv. sin \( \theta \) = \( \frac{1}{2} \) ..(i) [Given] In right angled \( \triangle \)ABC, ∠C = \( \theta \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC, जिसमें कोण C थीटा (θ) है, कोण B समकोण है, AC भुजा की लंबाई 2k है और AB भुजा की लंबाई 1k है। sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} \) sin \( \theta \) = \( \frac{AB}{AC} \) ...(ii)
\( \implies \frac{AB}{AC} = \frac{1}{2} \) ...[From (i) and (ii)] Let the common multiple be k. .: AB = 1k and AC = 2k Now, AC\(^2\) = AB\(^2\) + BC\(^2\) ...[Pythagoras theorem] .: (2k)\(^2\) = (1k)\(^2\) + BC\(^2\) .: 4k\(^2\) = k\(^2\) + BC\(^2\) .: BC\(^2\) = 4k\(^2\) - k\(^2\) = 3k\(^2\) .: BC = \( \sqrt{3k^2} \) ...[Taking square root of both sides] = \( \sqrt{3} \)k cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2} \) tan \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Adjacent side of } \theta} = \frac{AB}{BC} = \frac{1k}{\sqrt{3}k} = \frac{1}{\sqrt{3}} \) v. cos \( \theta \) = \( \frac{1}{\sqrt{3}} \) ..(i) [Given] In right angled \( \triangle \)ABC, ∠C = \( \theta \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC, जिसमें कोण C थीटा (θ) है, कोण B समकोण है, BC भुजा की लंबाई 1k है और AC भुजा की लंबाई \( \sqrt{3} \)k है। cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} \) cos \( \theta \) = \( \frac{BC}{AC} \) ...(ii)
\( \implies \frac{BC}{AC} = \frac{1}{\sqrt{3}} \) ...[From (i) and (ii)] Let the common multiple be k. .: BC = 1k and AC = \( \sqrt{3} \)k Now, AC\(^2\) = AB\(^2\) + BC\(^2\) ...[Pythagoras theorem] .: (\( \sqrt{3} \)k)\(^2\) = AB\(^2\) + (1k)\(^2\) .: 3k\(^2\) = AB\(^2\) + k\(^2\) AB\(^2\) = 3k\(^2\) - k\(^2\) = 2k\(^2\) .: AB = \( \sqrt{2k^2} \) ...[Taking square root of both sides] AB = \( \sqrt{2} \)k sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{\sqrt{2}k}{\sqrt{3}k} = \frac{\sqrt{2}}{\sqrt{3}} \) tan \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Adjacent side of } \theta} = \frac{AB}{BC} = \frac{\sqrt{2}k}{1k} = \sqrt{2} \) vi. cos \( \theta \) = \( \frac{21}{20} \) ...(i) [Given] In right angled \( \triangle \)ABC, ∠C = \( \theta \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC, जिसमें कोण C थीटा (θ) है, कोण B समकोण है, AC भुजा की लंबाई 21k है और BC भुजा की लंबाई 20k है। tan \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Adjacent side of } \theta} \) tan \( \theta \) = \( \frac{AB}{BC} \) ...(ii)
\( \implies \frac{AB}{BC} = \frac{21}{20} \) ...[From (i) and (ii)] Let the common multiple be k. .: AB = 21k and BC = 20k Now, AC\(^2\) = AB\(^2\) + BC\(^2\) ...[Pythagoras theorem] AC\(^2\) = (21k)\(^2\) + (20k)\(^2\) AC\(^2\) = 441k\(^2\) + 400k\(^2\) AC\(^2\) = 841k\(^2\) .: AC = \( \sqrt{841k^2} \) ...[Taking square root of both sides] = 29k .: sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{21k}{29k} = \frac{21}{29} \) cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{20k}{29k} = \frac{20}{29} \) vii. tan \( \theta \) = \( \frac{8}{15} \) ..(i) [Given] In right angled \( \triangle \)ABC, ∠C = \( \theta \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC, जिसमें कोण C थीटा (θ) है, कोण B समकोण है, AB भुजा की लंबाई 8k है और BC भुजा की लंबाई 15k है। tan \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Adjacent side of } \theta} \) tan \( \theta \) = \( \frac{AB}{BC} \) ...(ii)
\( \implies \frac{AB}{BC} = \frac{8}{15} \) ...[From (i) and (ii)] Let the common multiple be k. .: AB = 8k and BC = 15k Now, AC\(^2\) = AB\(^2\) + BC\(^2\) ...[Pythagoras theorem] AC\(^2\) = (8k)\(^2\) + (15k)\(^2\) AC\(^2\) = 64k\(^2\) + 225k\(^2\) AC\(^2\) = 289k\(^2\) .: AC = \( \sqrt{289k^2} \) ...[Taking square root of both sides] = 17k .: sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{8k}{17k} = \frac{8}{17} \) cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{15k}{17k} = \frac{15}{17} \) viii. sin \( \theta \) = \( \frac{3}{5} \) ..(i) [Given] In right angled \( \triangle \)ABC, ∠C = \( \theta \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC, जिसमें कोण C थीटा (θ) है, कोण B समकोण है, AB भुजा की लंबाई 3k है और AC भुजा की लंबाई 5k है। sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} \) .: sin \( \theta \) = \( \frac{AB}{AC} \) ...(ii)
\( \implies \frac{AB}{AC} = \frac{3}{5} \) ...[From (i) and (ii)] Let the common multiple be k. .: AB = 3k and AC = 5k Now, AC\(^2\) = AB\(^2\) + BC\(^2\) ...[Pythagoras theorem] .: (5k)\(^2\) = (3k)\(^2\) + BC\(^2\) .: 25k\(^2\) = 9k\(^2\) + BC\(^2\) .: BC\(^2\) = 25k\(^2\) - 9k\(^2\) = 16k\(^2\) .: BC = \( \sqrt{16k^2} \) ...[Taking square root of both sides] = 4k cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{4k}{5k} = \frac{4}{5} \) tan \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Adjacent side of } \theta} = \frac{AB}{BC} = \frac{3k}{4k} = \frac{3}{4} \) ix. tan \( \theta \) = \( \frac{1}{2\sqrt{2}} \) ..(i) [Given] In right angled \( \triangle \)ABC, ∠C = \( \theta \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC, जिसमें कोण C थीटा (θ) है, कोण B समकोण है, AB भुजा की लंबाई 1k है और BC भुजा की लंबाई \( 2\sqrt{2} \)k है। tan \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Adjacent side of } \theta} \) .: tan \( \theta \) = \( \frac{AB}{BC} \) ...(ii)
\( \implies \frac{AB}{BC} = \frac{1}{2\sqrt{2}} \) ...[From (i) and (ii)] Let the common multiple be k. .: AB = 1k and BC = \( 2\sqrt{2} \)k Now, AC\(^2\) = AB\(^2\) + BC\(^2\) ...[Pythagoras theorem] = (1k)\(^2\) + (\( 2\sqrt{2} \)k)\(^2\) = k\(^2\) + (4 \(\times\) 2)k\(^2\) = k\(^2\) + 8k\(^2\) = 9k\(^2\) .: AC = \( \sqrt{9k^2} \) ...[Taking square root of both sides] = 3k sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{1k}{3k} = \frac{1}{3} \) cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{2\sqrt{2}k}{3k} = \frac{2\sqrt{2}}{3} \) Completed Table:

In simple words: For each column where one trigonometric ratio is given, we used the Pythagorean theorem and definitions of sine, cosine, and tangent in a right-angled triangle to find the lengths of the sides and then calculated the missing ratios. The common multiple 'k' was used to maintain proportionality of side lengths.

🎯 Exam Tip: Mastering the relationship between sides of a right-angled triangle and trigonometric ratios, along with the Pythagorean theorem, is crucial for full marks in such table-completion questions. Always simplify ratios to their lowest terms.

Question 2. Find the values of:
i. 5 sin 30° + 3 tan 45°
ii. \( \frac{4}{5} \) tan² 60° + 3 sin² 60°
iii. 2 sin 30° + cos 0° + 3 sin 90°
iv. \( \frac{\text{tan 60°}}{\text{sin 60°+cos 60°}} \)
v. cos² 45° + sin² 30°
vi. cos 60° x cos 30° + sin 60° x sin 30°
Answer: Solution: i. sin 30° = \( \frac{1}{2} \) and tan 45° = 1 5 sin 30° + 3 tan 45° = 5 \( \left( \frac{1}{2} \right) \) + 3(1) = \( \frac{5}{2} \) + 3 = \( \frac{5+6}{2} \)
\( \implies \) 5 sin 30° + 3 tan 45° = \( \frac{11}{2} \) ii. \( \frac{4}{5} \) tan² 60° + 3 sin² 60° \( \frac{4}{5} \) tan² 60° + 3 sin² 60° = \( \frac{4}{5} \) (tan 60°)\(^2\) + 3 (sin 60°)\(^2\) = \( \frac{4}{5} (\sqrt{3})^2 \) + 3 \( \left( \frac{\sqrt{3}}{2} \right)^2 \) = \( \frac{4}{5} \times 3 \) + 3 \( \times \frac{3}{4} \) = \( \frac{12}{5} \) + \( \frac{9}{4} \) = \( \frac{48+45}{20} \) = \( \frac{93}{20} \)
\( \implies \frac{4}{5} \) tan² 60° + 3 sin² 60° = \( \frac{93}{20} \) iii. 2 sin 30° + cos 0° + 3 sin 90° 2 sin 30° + cos0° + 3 sin 90° = 2 \( \left( \frac{1}{2} \right) \) + 1 + 3(1) = 1 + 1 + 3
\( \implies \) 2 sin 30° + cos 0° + 3 sin 90° = 5 iv. \( \frac{\text{tan 60°}}{\text{sin 60°+cos 60°}} \) \( \frac{\text{tan 60°}}{\text{sin 60° + cos 60°}} = \frac{\sqrt{3}}{\frac{\sqrt{3}}{2} + \frac{1}{2}} \) = \( \frac{\sqrt{3}}{\frac{\sqrt{3}+1}{2}} \) = \( \sqrt{3} \times \frac{2}{\sqrt{3}+1} \)
\( \implies \frac{\text{tan 60°}}{\text{sin 60° + cos 60°}} = \frac{2\sqrt{3}}{\sqrt{3}+1} \) v. cos² 45° + sin² 30° cos 45° = \( \frac{1}{\sqrt{2}} \) and sin 30° = \( \frac{1}{2} \) cos² 45° + sin² 30° = (cos 45°)\(^2\) + (sin 30°)\(^2\) = \( \left( \frac{1}{\sqrt{2}} \right)^2 \) + \( \left( \frac{1}{2} \right)^2 \) = \( \frac{1}{2} \) + \( \frac{1}{4} \) = \( \frac{2+1}{4} \)
\( \implies \) cos² 45° + sin² 30° = \( \frac{3}{4} \) vi. cos 60° x cos 30° + sin 60° x sin 30° cos 60° = \( \frac{1}{2} \), cos 30° = \( \frac{\sqrt{3}}{2} \), sin 60° = \( \frac{\sqrt{3}}{2} \) and sin 30° = \( \frac{1}{2} \) cos 60° x cos 30° + sin 60° x sin 30° = \( \frac{1}{2} \times \frac{\sqrt{3}}{2} \) + \( \frac{\sqrt{3}}{2} \times \frac{1}{2} \) = \( \frac{\sqrt{3}}{4} \) + \( \frac{\sqrt{3}}{4} \) = \( \frac{\sqrt{3}+\sqrt{3}}{4} \) = \( \frac{2\sqrt{3}}{4} \)
\( \implies \) cos 60° x cos 30° + sin 60° x sin 30° = \( \frac{\sqrt{3}}{2} \) In simple words: We calculated the values of given trigonometric expressions by substituting the standard values of trigonometric ratios for specific angles (0°, 30°, 45°, 60°, 90°) and performing the arithmetic operations. Each sub-part demonstrated the application of these fundamental values.

🎯 Exam Tip: Memorizing the trigonometric values for standard angles (0°, 30°, 45°, 60°, 90°) is critical. Mistakes often occur in basic arithmetic or squaring terms. Double-check your calculations.

Question 3. If sin \( \theta \) = \( \frac{4}{5} \), then find cos \( \theta \).
Answer: Solution: sin \( \theta \) = \( \frac{4}{5} \) ...(i)[Given] In right angled \( \triangle \)ABC, ∠C = \( \theta \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC, जिसमें कोण C थीटा (θ) है, कोण B समकोण है, AB भुजा की लंबाई 4k है और AC भुजा की लंबाई 5k है। sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} \) .: sin \( \theta \) = \( \frac{AB}{AC} \) ...(ii)
\( \implies \frac{AB}{AC} = \frac{4}{5} \) ...[From (i) and (ii)] Let the common multiple be k. .: AB = 4k and AC = 5k Now, AC\(^2\) = AB\(^2\) + BC\(^2\) ...[Pythagoras theorem] .: (5k)\(^2\) = (4k)\(^2\) + BC\(^2\) .: 25k\(^2\) = 16k\(^2\) + BC\(^2\) .: BC\(^2\) = 25k\(^2\) - 16k\(^2\) = 9k\(^2\) .: BC = \( \sqrt{9k^2} \) ...[Taking square root of both sides] = 3k .: cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{3k}{5k} = \frac{3}{5} \) In simple words: Given sin θ, we used the ratio definition to determine the lengths of the opposite side and hypotenuse in terms of 'k'. Then, applying the Pythagorean theorem, we found the adjacent side and subsequently calculated cos θ.

🎯 Exam Tip: Always draw a right-angled triangle to visualize the given ratio. Using 'k' for common multiples helps avoid errors in calculating side lengths. The Pythagorean identity (sin²θ + cos²θ = 1) can also be used as an alternative method for verification or direct calculation.

Question 4. If cos \( \theta \) = \( \frac{15}{17} \), then find sin \( \theta \).
Answer: Solution: cos \( \theta \) = \( \frac{15}{17} \) ...(i)[Given] In right angled \( \triangle \)ABC, ∠C = \( \theta \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC, जिसमें कोण C थीटा (θ) है, कोण B समकोण है, BC भुजा की लंबाई 15k है और AC भुजा की लंबाई 17k है। cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} \) .: cos \( \theta \) = \( \frac{BC}{AC} \) ...(ii)
\( \implies \frac{BC}{AC} = \frac{15}{17} \) ...[From (i) and (ii)] Let the common multiple be k. .: BC = 15k and AC = 17k Now, AC\(^2\) = AB\(^2\) + BC\(^2\) ...[Pythagoras theorem] .: (17k)\(^2\) = AB\(^2\) + (15k)\(^2\) .: 289k\(^2\) = AB\(^2\) + 225k\(^2\) .: AB\(^2\) = 289k\(^2\) - 225k\(^2\) = 64k\(^2\) .: AB = \( \sqrt{64k^2} \) ...[Taking square root of both sides] = 8k sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{8k}{17k} = \frac{8}{17} \) In simple words: Given cos θ, we used its definition to find the adjacent side and hypotenuse in terms of 'k'. The Pythagorean theorem was then used to find the opposite side, which allowed us to calculate sin θ.

🎯 Exam Tip: Similar to finding cos θ from sin θ, always draw a right-angled triangle. Remember the relationship between the sides and the trigonometric ratios. Using k ensures correct proportional side lengths, even if not explicitly asked.

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.2 Intext Questions and Activities

Question 1. In right angled \( \triangle \)PQR, ∠Q = 90°. Therefore ∠P and ∠R are complementary angles of each other. Verify the following ratios.
i. sin \( \theta \) = cos (90 - \( \theta \))
ii. cos \( \theta \) = sin (90 - \( \theta \))
iii. sin 30° = cos (90° - 30°) = cos 60°
iv. cos 30° = sin (90° - 30°) = sin 60° (Textbook pg. no. 107)
Answer: Solution: In \( \triangle \)PQR, ∠Q = 90°, ∠P = \( \theta \) .: ∠R = 90 - \( \theta \)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज PQR, जिसमें कोण Q समकोण है, कोण P थीटा (θ) है और कोण R (90 - θ) है। i. sin \( \theta \) = cos (90 - \( \theta \)) sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} \) = \( \frac{QR}{PR} \) ...(i) cos (90 - \( \theta \)) = \( \frac{\text{Adjacent side of } (90 - \theta)}{\text{Hypotenuse}} \) = \( \frac{QR}{PR} \) ...(ii) .: sin \( \theta \) = cos (90 - \( \theta \)) ...[From (i) and (ii)] ii. cos \( \theta \) = sin (90 - \( \theta \)) cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} \) = \( \frac{PQ}{PR} \) ...(i) sin (90 - \( \theta \)) = \( \frac{\text{Opposite side of } (90 - \theta)}{\text{Hypotenuse}} \) = \( \frac{PQ}{PR} \) ...(ii) .: cos \( \theta \) = sin (90 - \( \theta \)) ...[From (i) and (ii)] iii. Let ∠P = \( \theta \) = 30° .: ∠R = 90° - 30° sin 30° = \( \frac{\text{Opposite side of 30°}}{\text{Hypotenuse}} \) = \( \frac{QR}{PR} \) ...(i) cos (90° - 30°) = \( \frac{\text{Adjacent side of (90° - 30°)}}{\text{Hypotenuse}} \) = \( \frac{QR}{PR} \) ...(ii) sin 30° = cos (90° - 30°) ... [From (i) and (ii)] sin 30° = cos 60° iv. cos 30° = sin (90° - 30°) = sin 60° cos 30° = \( \frac{\text{Adjacent side of 30°}}{\text{Hypotenuse}} \) ...[.: \( \theta \) = 30°] = \( \frac{PQ}{PR} \) ...(i) sin (90° - 30°) = \( \frac{\text{Opposite side of (90° - 30°)}}{\text{Hypotenuse}} \) = \( \frac{PQ}{PR} \) ...(ii) .: cos 30° = sin (90° - 30°) ...[From (i) and (ii)] .: cos 30° = sin 60° In simple words: This activity verifies the complementary angle identities for sine and cosine (sin θ = cos(90°-θ) and cos θ = sin(90°-θ)) by using the definitions of these ratios in a right-angled triangle PQR, where P and R are complementary angles.

🎯 Exam Tip: Understanding complementary angle identities is fundamental for trigonometry. Ensure you can clearly define sine and cosine in terms of opposite, adjacent, and hypotenuse with respect to different angles in the same triangle. This concept is frequently tested.

Question 2. In right angled \( \triangle \)PQR, ∠Q = 90°, ∠R = \( \theta \) and if sin \( \theta \) = \( \frac{5}{13} \), then find cos \( \theta \) and tan \( \theta \). (Textbook pg. no. 110)
Answer: Solution: i. Take the given trigonometric ratio as 13k equation (i). sin \( \theta \) = \( \frac{5}{13} \) ...(i)[Given] By using the definition write the trigonometric ratio of sin \( \theta \) and take it as equation (ii). In right angled \( \triangle \)PQR, ∠R = \( \theta \)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज PQR, जिसमें कोण Q समकोण है, कोण R थीटा (θ) है, PR भुजा की लंबाई 13k है और PQ भुजा की लंबाई 5k है। sin \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} \) sin \( \theta \) = \( \frac{PQ}{PR} \) ...(ii)
\( \implies \frac{PQ}{PR} = \frac{5}{13} \) ...[From (i) and (ii)] Let the common multiple be k. .: PQ = 5k and PR = 13k Find QR by using Pythagoras theorem. PR\(^2\) = PQ\(^2\) + QR\(^2\) ...[Pythagoras theorem] .: (13k)\(^2\) = (5k)\(^2\) + QR\(^2\) .: 169k\(^2\) = 25k\(^2\) + QR\(^2\) .: QR\(^2\) = 169k\(^2\) - 25k\(^2\) = 144k\(^2\) .: QR = \( \sqrt{144k^2} \) ...[Taking square root of both sides] = 12k cos \( \theta \) = \( \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{12k}{13k} = \frac{12}{13} \) tan \( \theta \) = \( \frac{\text{Opposite side of } \theta}{\text{Adjacent side of } \theta} = \frac{PQ}{QR} = \frac{5k}{12k} = \frac{5}{12} \) In simple words: Given sin θ in a right-angled triangle, we identified the opposite side and hypotenuse. Using the Pythagorean theorem, we calculated the adjacent side. With all three side lengths, we then found the values for cos θ and tan θ.

🎯 Exam Tip: This question tests your ability to apply basic trigonometric definitions and the Pythagorean theorem. Ensure you correctly identify the opposite, adjacent, and hypotenuse sides relative to the given angle. Using 'k' as a common multiple is a good practice to represent side lengths accurately.

Question 3. While solving the above Illustrative example, why the lengths of PQ and PR are taken 5k and 13k? (Textbook pg. no. 111)
Answer: Solution: \( \frac{PQ}{PR} = \frac{5}{13} \) ...[Given] Here, the ratio of the lengths of sides PQ and PR is 5 : 13. The actual lengths of the sides can be any multiple of the ratio. Hence, we consider the multiple k while solving. In simple words: Side lengths are taken as 5k and 13k because the given values (5 and 13) represent a ratio, not the actual lengths. 'k' is a common multiple that scales the ratio to the true side lengths, allowing the Pythagorean theorem to be applied correctly.

🎯 Exam Tip: Always remember that a trigonometric ratio like \( \frac{5}{13} \) represents a proportion, not absolute lengths. Introducing a common multiple 'k' is essential to correctly calculate the third side using the Pythagorean theorem and avoid errors that would arise from assuming k=1 prematurely.

Question 4. While solving the above illustrative example, can we take the lengths of PQ and PR as 5 and 13? If so, then what changes are needed In the writing of the solution. (Textbook pg. no. 111)
Answer: Solution: Yes, we can take lengths of PQ and PR as 5 and 13. In that case, we will have to take k = 1 and solve the problem accordingly. In simple words: Yes, we can directly use 5 and 13 as side lengths if we assume the common multiple 'k' to be 1. The solution would then proceed by replacing all instances of 'k' with 1.

🎯 Exam Tip: While assuming k=1 can simplify calculations, it's crucial to understand that you are making an assumption. For questions where the actual side lengths might be different, explicitly using 'k' is more rigorous. If the question doesn't require actual lengths, k=1 is often implied.

Question 5. Verify that the equation 'sin² \( \theta \) + cos² \( \theta \) = 1′ is true when \( \theta \) = 0° or \( \theta \) = 90°. (Textbook pg. no. 112)
Answer: Solution: sin² \( \theta \) + cos² \( \theta \) = 1 i. If \( \theta \) = 0°, L.H.S. = sin² \( \theta \) + cos² \( \theta \) = sin² 0° + cos² 0° = 0 + 1 ...[.: sin 0° = 0, cos 0° = 1] = R.H.S. .: sin² \( \theta \) + cos² \( \theta \) = 1 ii. If \( \theta \) = 90°, L.H.S.= sin² \( \theta \) +cos² \( \theta \) = sin² 90° + cos² 90° = 1 + 0 ... [ sin 90° = 1, cos 90° = 0] = 1 = R.H.S. .: sin² \( \theta \) + cos² \( \theta \) = 1 In simple words: We verified the fundamental trigonometric identity sin²θ + cos²θ = 1 by substituting the known values of sine and cosine for angles 0° and 90°. In both cases, the sum of the squares of sine and cosine equaled 1, thus proving the identity.

🎯 Exam Tip: The identity sin²θ + cos²θ = 1 is fundamental. Knowing the standard trigonometric values for 0° and 90° is essential for quickly verifying such identities. Pay attention to squaring the values correctly.

Question 3. While solving the above Illustrative example, why the lengths of PQ and PR are taken 5k and 13k? (Textbook pg. no. 111)
Answer: PQ/PR = 5/13... [Given]
Here, the ratio of the lengths of sides PQ and PR is 5 : 13.
The actual lengths of the sides can be any multiple of the ratio. Hence, we consider the multiple k while solving.
In simple words: The ratio 5:13 represents a proportional relationship between the side lengths. To express their actual lengths while maintaining this proportion, we introduce a common multiple 'k'.

🎯 Exam Tip: Understanding the role of the common multiple 'k' in trigonometric ratios is crucial for solving problems involving unknown side lengths accurately.

Question 4. While solving the above illustrative example, can we take the lengths of PQ and PR as 5 and 13? If so, then what changes are needed In the writing of the solution. (Tcxtbook pg. no. 111)
Answer: Yes, we can take lengths of PQ and PR as 5 and 13.
In that case, we will have to take k = 1 and solve the problem accordingly.
In simple words: Yes, you can use 5 and 13 directly if you assume 'k' is 1, simplifying the calculation while still preserving the given ratio.

🎯 Exam Tip: When the common multiple 'k' is 1, the numerical values of the ratio directly represent the side lengths, streamlining the solution process.

Question 5. Verify that the equation 'sin² \( \theta \) + cos² \( \theta \) = 1′ is true when \( \theta \) = 0° or \( \theta \) = 90°. (Textbook pg. no. 112)
Answer: sin² \( \theta \) + cos² \( \theta \) = 1
(i) If \( \theta \) = 0°,
LH.S. = sin² \( \theta \) + cos² \( \theta \)
= sin² 0° + cos² 0°
= 0 + 1 ...[\(\therefore\) sin 0° = 0, cos 0° = 1]
= R.H.S.
\(\therefore\) sin² \( \theta \) + cos² \( \theta \) = 1
(ii) If \( \theta \) = 90°,
L.H.S.= sin² \( \theta \) +cos² \( \theta \)
= sin² 90° + cos² 90°
= 1 + 0 ... [\(\therefore\) sin 90° = 1, cos 90° = 0]
= 1
= R.H.S.
\(\therefore\) sin² \( \theta \) + cos² \( \theta \) = 1
In simple words: The fundamental trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \) holds true for both \( \theta \) = 0° and \( \theta \) = 90° as verified by substituting the respective standard angle values.

🎯 Exam Tip: This identity is fundamental in trigonometry; remember the standard values for sine and cosine at 0° and 90° to quickly verify or apply it in problems.