Chapter 8 — Trigonometry — Quick Q&A & Textbook Solutions

20 Most Important — 1 Mark Questions

Q1. What is trigonometry?
A1. Trigonometry studies relations between angles and sides of triangles, especially right-angled triangles.
Q2. Define sine of an acute angle in a right triangle.
A2. \(\sin \theta = \dfrac{\text{opposite side}}{\text{hypotenuse}}\).
Q3. Define cosine of an acute angle.
A3. \(\cos \theta = \dfrac{\text{adjacent side}}{\text{hypotenuse}}\).
Q4. Define tangent of an acute angle.
A4. \(\tan \theta = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{\sin\theta}{\cos\theta}\).
Q5. What is the hypotenuse?
A5. The side opposite the right angle — the longest side of a right triangle.
Q6. Write the identity relating \(\sin^2\theta\) and \(\cos^2\theta\).
A6. \(\sin^2\theta+\cos^2\theta=1\).
Q7. What is \(\sin 30^\circ\)?
A7. \(\sin 30^\circ = \dfrac{1}{2}\).
Q8. What is \(\cos 60^\circ\)?
A8. \(\cos 60^\circ = \dfrac{1}{2}\).
Q9. What is \(\tan 45^\circ\)?
A9. \(\tan 45^\circ = 1\).
Q10. What is \(\sin 90^\circ\)?
A10. \(\sin 90^\circ = 1\).
Q11. What is \(\cos 0^\circ\)?
A11. \(\cos 0^\circ = 1\).
Q12. What is \(\tan 0^\circ\)?
A12. \(\tan 0^\circ = 0\).
Q13. State complementary-angle relation for sine and cosine.
A13. \(\sin\theta=\cos(90^\circ-\theta)\) and \(\cos\theta=\sin(90^\circ-\theta)\).
Q14. What is \(\tan 60^\circ\)?
A14. \(\tan 60^\circ = \sqrt{3}\) (often written \(=\dfrac{\sqrt3}{1}\)).
Q15. What is the tangent in terms of sine and cosine?
A15. \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) (provided \(\cos\theta\neq0\)).
Q16. What is \(\sin 45^\circ\)?
A16. \(\sin 45^\circ = \dfrac{\sqrt2}{2}\) (same as \(\cos45^\circ\)).
Q17. In a right triangle, which two acute angles are complementary?
A17. The two non-right angles (they add up to \(90^\circ\)).
Q18. What is \(\sin 0^\circ\)?
A18. \(\sin 0^\circ = 0\).
Q19. Is \(\tan 90^\circ\) defined?
A19. No — \(\tan 90^\circ\) is undefined (division by zero since \(\cos90^\circ=0\)).
Q20. What are the reciprocal (inverse) trig ratios of \(\sin,\cos,\tan\)?
A20. \(\csc\theta=\dfrac{1}{\sin\theta},\ \sec\theta=\dfrac{1}{\cos\theta},\ \cot\theta=\dfrac{1}{\tan\theta}.\)

20 Most Important — 2 Marks Questions

Q1. In right \(\triangle ABC\) with \(\angle B=90^\circ\), opposite \(A\) is \(BC\), adjacent to \(A\) is \(AB\), write \(\sin A,\cos A,\tan A\).
A1. \(\sin A=\dfrac{BC}{AC},\ \cos A=\dfrac{AB}{AC},\ \tan A=\dfrac{BC}{AB}.\)
Q2. Express \(\sin(90^\circ-\theta)\) in terms of sine or cosine of \(\theta\).
A2. \(\sin(90^\circ-\theta)=\cos\theta.\)
Q3. If \(\sin\theta=\dfrac{3}{5}\) and \(\theta\) acute, find \(\cos\theta\) and \(\tan\theta\).
A3. \(\cos\theta=\sqrt{1-\frac{9}{25}}=\frac{4}{5},\ \tan\theta=\dfrac{3/5}{4/5}=\dfrac{3}{4}.\)
Q4. If \(\cos\theta=\dfrac{5}{13}\) (acute), find \(\sin\theta,\tan\theta\).
A4. \(\sin\theta=\sqrt{1-\frac{25}{169}}=\frac{12}{13},\ \tan\theta=\dfrac{12}{5}.\)
Q5. Evaluate \(2\tan45^\circ+\cos30^\circ-\sin60^\circ\).
A5. \(2(1)+\dfrac{\sqrt3}{2}-\dfrac{\sqrt3}{2}=2.\)
Q6. Using 30°–60°–90° triangle, write \(\sin60^\circ,\cos60^\circ,\tan30^\circ.\)
A6. \(\sin60^\circ=\dfrac{\sqrt3}{2},\ \cos60^\circ=\dfrac12,\ \tan30^\circ=\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}.\)
Q7. Evaluate \(\sin34^\circ\) vs \(\cos56^\circ\).
A7. \(\sin34^\circ=\cos(90^\circ-34^\circ)=\cos56^\circ\) (they are equal).
Q8. In right \(\triangle PQR\) with \(\angle Q=90^\circ\), if \(\sin R=\dfrac{5}{13}\), find \(\cos R,\tan R\).
A8. \(\cos R=\dfrac{12}{13},\ \tan R=\dfrac{5}{12}\) (from Pythagorean triple \(5,12,13\)).
Q9. Prove \(\tan\theta\cdot\tan(90^\circ-\theta)=1\).
A9. \(\tan(90^\circ-\theta)=\cot\theta=\dfrac{1}{\tan\theta}\). Hence product \(=1\).
Q10. Give values: \(\sin45^\circ,\cos45^\circ,\tan45^\circ\).
A10. \(\sin45^\circ=\cos45^\circ=\dfrac{\sqrt2}{2},\ \tan45^\circ=1.\)
Q11. If \(\sin\theta=\dfrac{4}{5}\), find \(\csc\theta\).
A11. \(\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{5}{4}.\)
Q12. Simplify \(\sin^2\theta+\cos^2\theta\).
A12. \(\sin^2\theta+\cos^2\theta=1.\)
Q13. Using 45°–45°–90° triangle with legs \(a\), hypotenuse \(a\sqrt2\), show \(\sin45^\circ=\dfrac{1}{\sqrt2}\).
A13. \(\sin45^\circ=\dfrac{\text{opposite}}{\text{hyp}}=\dfrac{a}{a\sqrt2}=\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}.\)
Q14. Find \(\tan(90^\circ-\theta)\) in terms of \(\theta\).
A14. \(\tan(90^\circ-\theta)=\cot\theta=\dfrac{1}{\tan\theta}.\)
Q15. Evaluate \(\sin0^\circ+\sin90^\circ\).
A15. \(0+1=1.\)
Q16. If \(\tan\theta=1\), what are possible acute \(\theta\)?
A16. \(\theta=45^\circ\) (for acute angle between 0° and 90°).
Q17. If \(\cos\theta=\sin(90^\circ-\theta)\), verify numerically for \(\theta=30^\circ\).
A17. \(\cos30^\circ=\dfrac{\sqrt3}{2}\). \(\sin60^\circ=\dfrac{\sqrt3}{2}\). They match.
Q18. Show \(\sin30^\circ=\cos60^\circ\).
A18. \(\sin30^\circ=\dfrac{1}{2},\ \cos60^\circ=\dfrac{1}{2}.\)
Q19. If \(\sin\theta=\dfrac{7}{25}\) with \(\theta\) acute, find \(\cos\theta\) and \(\tan\theta\).
A19. \(\cos\theta=\sqrt{1-\frac{49}{625}}=\sqrt{\frac{576}{625}}=\dfrac{24}{25},\ \tan\theta=\dfrac{7}{24}.\)
Q20. Convert \(\sin 34^\circ=\cos56^\circ\) to complementary-angle statement.
A20. Since \(34^\circ+56^\circ=90^\circ\), \(\sin34^\circ=\cos(90^\circ-34^\circ)=\cos56^\circ\).

20 Most Important — 3 Marks Questions

Q1. Prove \(\sin^2\theta+\cos^2\theta=1\) using a right triangle.
A1. Let right triangle have hypotenuse \(h\), opposite \(o\), adjacent \(a\). Then \(\sin\theta=\dfrac{o}{h},\ \cos\theta=\dfrac{a}{h}\). By Pythagoras \(o^2+a^2=h^2\). Divide by \(h^2\): \(\left(\dfrac{o}{h}\right)^2+\left(\dfrac{a}{h}\right)^2=1\Rightarrow\sin^2\theta+\cos^2\theta=1\).
Q2. In a triangle with sides 3,4,5 (right triangle), find \(\sin A,\cos A,\tan B\) where angle opposite side 3 is A.
A2. Hypotenuse = 5. If side opposite A = 3 then adjacent = 4. \(\sin A=\dfrac{3}{5},\ \cos A=\dfrac{4}{5}\). Angle B opposite 4: \(\tan B=\dfrac{4}{3}?\) Wait: if B opposite 4 then \(\tan B=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{4}{3}.\)
Q3. Using 30°–60°–90° triangle derive \(\sin30^\circ,\cos30^\circ,\tan60^\circ\).
A3. For sides \(1:\sqrt3:2\) (opposite 30°, opposite 60°, hypotenuse 2): \(\sin30^\circ=\dfrac{1}{2},\ \cos30^\circ=\dfrac{\sqrt3}{2},\ \tan60^\circ=\dfrac{\sqrt3}{1}=\sqrt3.\)
Q4. If \(\sin\theta=\dfrac{5}{13}\), find \(\cos\theta,\tan\theta,\sec\theta\).
A4. \(\cos\theta=\dfrac{12}{13},\ \tan\theta=\dfrac{5}{12},\ \sec\theta=\dfrac{13}{12}.\)
Q5. Evaluate \(2\sin45^\circ + \cos60^\circ - \tan30^\circ\).
A5. \(2\cdot\frac{\sqrt2}{2} + \frac12 - \frac{1}{\sqrt3} = \sqrt2 + \frac12 - \frac{\sqrt3}{3}.\) (Exact simplified form.)
Q6. If \(\cos\theta=\dfrac{24}{25}\) (acute), compute \(\sin^2\theta\) and \(\tan\theta\).
A6. \(\sin^2\theta = 1-\cos^2\theta=1-\left(\tfrac{24}{25}\right)^2=1-\tfrac{576}{625}=\tfrac{49}{625}\Rightarrow\sin\theta=\tfrac{7}{25}.\ \tan\theta=\tfrac{7}{24}.\)
Q7. Prove \(\sin(90^\circ-\theta)=\cos\theta\) using triangle.
A7. In a right triangle the non-right angles are complementary. For angle \(\theta\), sine equals opposite/hypotenuse, which is same as cosine of the complementary angle. So \(\sin(90^\circ-\theta)=\cos\theta\).
Q8. Solve: In right triangle, if leg lengths are 8 and 15, find all six trig ratios for angle opposite 8.
A8. Hypotenuse = \(\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17\). \(\sin=\tfrac{8}{17},\cos=\tfrac{15}{17},\tan=\tfrac{8}{15},\csc=\tfrac{17}{8},\sec=\tfrac{17}{15},\cot=\tfrac{15}{8}.\)
Q9. Explain why \(\tan 90^\circ\) is undefined.
A9. \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\). At \(90^\circ\), \(\cos90^\circ=0\), division by zero is undefined, so \(\tan90^\circ\) is undefined.
Q10. Using complementary relations, express \(\cot\theta\) as a trig ratio of complement.
A10. \(\cot\theta=\tan(90^\circ-\theta)\).
Q11. If \(\sin\alpha= \tfrac{3}{5}\) and \(\sin\beta=\tfrac{5}{13}\), find \(\sin\alpha\cos\beta + \cos\alpha\sin\beta\) (use identity).
A11. This equals \(\sin(\alpha+\beta)\). Compute cosines: \(\cos\alpha=\tfrac{4}{5},\cos\beta=\tfrac{12}{13}\). So value \(=\tfrac{3}{5}\cdot\tfrac{12}{13}+\tfrac{4}{5}\cdot\tfrac{5}{13}=\tfrac{36+20}{65}=\tfrac{56}{65}.\)
Q12. Prove \(\sin30^\circ=\frac12\) using geometry of 30-60-90 triangle.
A12. In equilateral triangle side \(2a\), drop altitude → two right triangles with hypotenuse \(2a\), short leg \(a\). For 30° angle opposite short leg: \(\sin30^\circ=\dfrac{a}{2a}=\tfrac12.\)
Q13. If \(\tan\theta=\dfrac{5}{12}\) and \(\theta\) acute, find \(\sin\theta\) and \(\cos\theta\).
A13. Let opposite=5k, adjacent=12k → hypotenuse \(13k\). \(\sin\theta=\tfrac{5}{13},\ \cos\theta=\tfrac{12}{13}.\)
Q14. Show that \(\sin(30^\circ)=\cos(60^\circ)\) by value substitution.
A14. \(\sin30^\circ=\tfrac12,\ \cos60^\circ=\tfrac12\) ⇒ equal.
Q15. Compute \(\sin^2 45^\circ + \cos^2 45^\circ\).
A15. Each component = \(\left(\tfrac{\sqrt2}{2}\right)^2=\tfrac12\). Sum \(=\tfrac12+\tfrac12=1.\)
Q16. If in triangle \(PQR\) with \(\angle Q=90^\circ\), \(PQ=5,PR=13\), find \(QR\) and trig ratios of angle \(R\) (opposite PQ).
A16. By Pythagoras \(QR=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\). For \(\angle R\): \(\sin R=\dfrac{5}{13},\cos R=\dfrac{12}{13},\tan R=\dfrac{5}{12}.\)
Q17. Evaluate \(\sin56^\circ\) vs \(\cos34^\circ\).
A17. Since \(56^\circ+34^\circ=90^\circ\), \(\sin56^\circ=\cos34^\circ.\)
Q18. If \(\sin\theta=0\), what are possible \(\theta\) (in degrees within \(0^\circ\)–\(360^\circ\))?
A18. \(\theta=0^\circ,180^\circ,360^\circ\) (general solution \(\theta=n\cdot180^\circ\)).
Q19. Show that \(\tan30^\circ\cdot\tan60^\circ=1\).
A19. \(\tan30^\circ=\dfrac{1}{\sqrt3},\ \tan60^\circ=\sqrt3\). Product \(=1.\)
Q20. For small practical measurement: How does trigonometry help find height of a tree?
A20. Measure shadow length and angle of elevation (or use similar triangles with a stick + its shadow) then use trig ratio: height = shadow×tan(angle) or use similarity proportion.

Textbook Exercises — Perfect Solutions (Practice Set 8.1, 8.2 & Problem Set 8)

Practice 8.1 — Q1 (Fig. 8.12). In \(\triangle PQR\) with \(\angle R=90^\circ\) write: (i) \(\sin P\) (ii) \(\cos Q\) (iii) \(\tan P\) (iv) \(\tan Q\).
A1. Hypotenuse = \(PQ\). Opposite to \(P\) is \(QR\); adjacent to \(P\) is \(PR\).
\(\sin P=\dfrac{QR}{PQ},\quad \cos Q=\dfrac{QR}{PQ},\quad \tan P=\dfrac{QR}{PR},\quad \tan Q=\dfrac{PR}{QR}.\)
Practice 8.1 — Q2 (Fig.8.13). In right \(\triangle XYZ\) (right at Y) with sides labelled \(a,b,c\) find: (i) \(\sin X\) (ii) \(\tan Z\) (iii) \(\cos X\) (iv) \(\tan X\).
A2. Interpret sides: take \(XZ=c\) as hypotenuse, opposite \(X\) is \(YZ=b\), adjacent to \(X\) is \(XY=a\).
\(\sin X=\dfrac{b}{c},\quad \tan Z=\dfrac{b}{a},\quad \cos X=\dfrac{a}{c},\quad \tan X=\dfrac{b}{a}.\)
Practice 8.1 — Q3 (Fig.8.14). Right \(\triangle LMN\) with \(\angle L=50^\circ,\angle N=40^\circ\). Write: (i) \(\sin50^\circ\) (ii) \(\cos50^\circ\) (iii) \(\tan40^\circ\) (iv) \(\cos40^\circ\).
A3. These are direct values/symbols — write them as functions:
\(\sin50^\circ,\ \cos50^\circ,\ \tan40^\circ,\ \cos40^\circ.\)
(If numerical values required use calculator; textbook expects the expression form.)
Practice 8.1 — Q4 (Fig.8.15). Two right triangles share a common right angle setup. Write (i) \(\sin a,\cos a,\tan a\) and (ii) \(\sin q,\cos q,\tan q\).
A4. Using triangle labelling: for angle \(a\): \(\sin a=\dfrac{\text{opp to }a}{\text{hyp to }a},\ \cos a=\dfrac{\text{adj to }a}{\text{hyp to }a},\ \tan a=\dfrac{\text{opp}}{\text{adj}}.\)
Similarly for \(q\). (Textbook problem asks to identify correct opposite/adjacent/hypotenuse from diagram — the pattern is the same.)
Solved Example — Ex(1). Find value of \(2\tan45^\circ+\cos30^\circ-\sin60^\circ\).
Solution. \(\tan45^\circ=1,\ \cos30^\circ=\dfrac{\sqrt3}{2},\ \sin60^\circ=\dfrac{\sqrt3}{2}\). So value \(=2\cdot1+\dfrac{\sqrt3}{2}-\dfrac{\sqrt3}{2}=2.\)
Solved Example — Ex(3). In right \(\triangle ACB\) with \(\angle C=90^\circ,\ AC=3,BC=4\). Find \(\sin A,\sin B,\cos A,\tan B\).
Solution. Hypotenuse \(AB=\sqrt{3^2+4^2}=\sqrt{25}=5\). \(\sin A=\dfrac{BC}{AB}=\dfrac{4}{5},\ \cos A=\dfrac{AC}{AB}=\dfrac{3}{5},\ \sin B=\dfrac{AC}{AB}=\dfrac{3}{5},\ \tan B=\dfrac{AC}{BC}=\dfrac{3}{4}.\)
Solved Example — Ex(4). In right \(\triangle PQR\) with \(\angle Q=90^\circ,\ \angle R=\theta,\ \sin\theta=\dfrac{5}{13}\). Find \(\cos\theta,\tan\theta\).
Solution. Take sides \(PQ=5k, PR=13k\Rightarrow QR=\sqrt{13^2-5^2}\,k=12k\). \(\cos\theta=\dfrac{QR}{PR}=\dfrac{12}{13},\ \tan\theta=\dfrac{PQ}{QR}=\dfrac{5}{12}.\)
Practice 8.2 — Q1 (table). If one trig ratio is given find the other two (examples).
A1. Example solutions (method & examples):
— If \(\sin\theta=\dfrac{11}{61}\) then \(\cos\theta=\sqrt{1-\left(\tfrac{11}{61}\right)^2}=\dfrac{\sqrt{61^2-11^2}}{61}=\dfrac{\sqrt{3721-121}}{61}=\dfrac{\sqrt{3600}}{61}=\dfrac{60}{61}.\ \tan\theta=\dfrac{11}{60}.\)
— If \(\cos\theta=\dfrac{3}{5}\) then \(\sin\theta=\dfrac{4}{5},\ \tan\theta=\dfrac{4}{3}.\)
(Use same method for other columns: compute missing sides by Pythagoras and form ratios.)
Practice 8.2 — Q2 Evaluate several expressions (interpreted reasonably):
A2. Interpreting items as typical textbook expressions:
(i) \(5\sin30^\circ + 3\tan45^\circ = 5\cdot\dfrac12 + 3\cdot1 = \dfrac{5}{2}+3=\dfrac{11}{2}=5.5.\)
(ii) \( \; \text{(example)}\ 4\tan60^\circ + 3\sin60^\circ = 4\sqrt3 + 3\cdot\dfrac{\sqrt3}{2}=4\sqrt3 + \dfrac{3\sqrt3}{2}=\dfrac{11\sqrt3}{2}.\)
(iii) \(\sin30^\circ+\cos90^\circ+\sin0^\circ=\tfrac12+0+0=\tfrac12.\)
(iv) \(\dfrac{\tan60^\circ}{\sin60^\circ+\cos60^\circ}=\dfrac{\sqrt3}{\frac{\sqrt3}{2}+\frac12}=\dfrac{\sqrt3}{\frac{\sqrt3+1}{2}}=\dfrac{2\sqrt3}{\sqrt3+1}.\)
(v) \(\cos^2 45^\circ + \sin^2 30^\circ = \left(\tfrac{\sqrt2}{2}\right)^2 + \left(\tfrac12\right)^2 = \tfrac12+\tfrac14=\tfrac34.\)
(vi) \(\cos60^\circ\cos30^\circ + \sin60^\circ\sin30^\circ = \left(\tfrac12\cdot\tfrac{\sqrt3}{2}\right)+\left(\tfrac{\sqrt3}{2}\cdot\tfrac12\right)=\tfrac{\sqrt3}{4}+\tfrac{\sqrt3}{4}=\tfrac{\sqrt3}{2}.\)
**Note:** Some textbook lines were garbled in the scanned text; above I interpreted the most likely standard expressions and solved them. If you want exact column-by-column solutions from the original table, paste that section as plain text and I'll fill it precisely.
Practice 8.2 — Q3. If \(\sin\theta=\dfrac{4}{5}\), find \(\cos\theta\).
A3. \(\cos\theta=\sqrt{1-\left(\tfrac{4}{5}\right)^2}=\sqrt{1-\tfrac{16}{25}}=\sqrt{\tfrac{9}{25}}=\tfrac{3}{5}.\)
Practice 8.2 — Q4. If \(\cos\theta=\dfrac{15}{17}\), find \(\sin\theta\).
A4. \(\sin\theta=\sqrt{1-\left(\tfrac{15}{17}\right)^2}=\sqrt{\tfrac{289-225}{289}}=\sqrt{\tfrac{64}{289}}=\tfrac{8}{17}.\)
Problem Set 8 — Q1 (MCQ). (i) Which statement true? \(\sin\theta=\cos(90^\circ-\theta)\) etc.
A1(i). True statement: \(\sin\theta=\cos(90^\circ-\theta)\). Answer: (A).
Problem Set 8 — Q1 (ii). Value of \(\sin90^\circ\)?
A1(ii). \(\sin90^\circ=1.\) Answer: (D).
Problem Set 8 — Q1 (iii). Evaluate \(2\tan45^\circ+\cos45^\circ-\sin45^\circ\).
A1(iii). \(2\cdot1+\tfrac{\sqrt2}{2}-\tfrac{\sqrt2}{2}=2.\) Answer: (C) 2.
Problem Set 8 — Q1 (iv). \(\dfrac{\cos28^\circ}{\sin62^\circ} = ?\)
A1(iv). Since \(62^\circ=90^\circ-28^\circ\), \(\sin62^\circ=\cos28^\circ\). Ratio \(=1\). Answer: (D).
Problem Set 8 — Q2. In right \(\triangle TSU\) with \(TS=5,SU=12,\angle S=90^\circ\) find \(\sin T,\cos T,\tan T\) and for \(U\).
A2. Hypotenuse \(TU=\sqrt{5^2+12^2}=13\). For angle \(T\): \(\sin T=\dfrac{12}{13},\ \cos T=\dfrac{5}{13},\ \tan T=\dfrac{12}{5}.\) For \(U\): \(\sin U=\dfrac{5}{13},\ \cos U=\dfrac{12}{13},\ \tan U=\dfrac{5}{12}.\)
Problem Set 8 — Q3. In right \(\triangle YXZ\) with \(X=90^\circ, XZ=8, YZ=17\). Find all six trig ratios for \(Y\) and \(Z\).
A3. Other leg \(XY=\sqrt{17^2-8^2}=\sqrt{289-64}=15.\) So for \(Y\): \(\sin Y=\dfrac{8}{17},\ \cos Y=\dfrac{15}{17},\ \tan Y=\dfrac{8}{15}.\) For \(Z\): \(\sin Z=\dfrac{15}{17},\ \cos Z=\dfrac{8}{17},\ \tan Z=\dfrac{15}{8}.\)
Problem Set 8 — Q4. In right \(\triangle LMN\) with \(\angle M=90^\circ,\ \cos q=\dfrac{24}{25}\). Find \(\sin q,\tan q,\sin^2 q,\cos^2 q\).
A4. \(\sin q=\sqrt{1-(24/25)^2}=\dfrac{7}{25}.\ \tan q=\dfrac{7}{24}.\ \sin^2 q=\dfrac{49}{625},\ \cos^2 q=\dfrac{576}{625}.\)
Problem Set 8 — Q5 (Fill blanks): (i) \(\sin20^\circ=\cos\_\_\_\) (ii) \(\tan30^\circ\cdot\tan\_\_\_ =1\) (iii) \(\cos40^\circ=\sin\_\_\_\)
A5. (i) \(\sin20^\circ=\cos70^\circ\). (ii) \(\tan30^\circ\cdot\tan60^\circ=1\). (iii) \(\cos40^\circ=\sin50^\circ.\)

Notes about ambiguous scanned lines: a few lines in the pasted textbook text were garbled (tables / expressions). I made the most likely interpretations (standard textbook exercises) and solved them precisely. If you want exact cell-by-cell filling from the original table image/text, paste that table as clean text and I will update the answers exactly.

Quick Reference & Useful Trig Values

Q. Key trig values for common angles
\[ \begin{aligned} &\sin0^\circ=0,\ \sin30^\circ=\tfrac12,\ \sin45^\circ=\tfrac{\sqrt2}{2},\ \sin60^\circ=\tfrac{\sqrt3}{2},\ \sin90^\circ=1,\\ &\cos0^\circ=1,\ \cos30^\circ=\tfrac{\sqrt3}{2},\ \cos45^\circ=\tfrac{\sqrt2}{2},\ \cos60^\circ=\tfrac12,\ \cos90^\circ=0,\\ &\tan0^\circ=0,\ \tan30^\circ=\tfrac{1}{\sqrt3},\ \tan45^\circ=1,\ \tan60^\circ=\sqrt3. \end{aligned} \]
Q. Important identities
\(\sin^2\theta+\cos^2\theta=1,\ \tan\theta=\dfrac{\sin\theta}{\cos\theta},\ \sin(90^\circ-\theta)=\cos\theta.\)
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