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NCERT Solutions · Class 10 · Chapter 8: Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1

Exercise 8.1: Trigonometric ratios. In a right triangle, for acute angle A: \(\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}}\), \(\cos A = \dfrac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan A = \dfrac{\text{opposite}}{\text{adjacent}}\); cosec, sec, cot are their reciprocals. Given one ratio, draw the triangle and find the third side by Pythagoras.

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Exercise 8.1 questions and solutions

Exercise 8.1, Question 1

In \(\triangle ABC\), right-angled at B, \(AB = 24\) cm and \(BC = 7\) cm. Find:
(i) \(\sin A,\ \cos A\)
Show solution
  1. \(AC = \sqrt{24^2 + 7^2} = 25\).
  2. For A: opposite \(BC = 7\), adjacent \(AB = 24\).
Answer: \[\begin{gathered}\sin A = \tfrac{7}{25}, \\ \cos A = \tfrac{24}{25}\end{gathered}\]
(ii) \(\sin C,\ \cos C\)
Show solution
  1. For C: opposite \(AB = 24\), adjacent \(BC = 7\).
Answer: \[\begin{gathered}\sin C = \tfrac{24}{25}, \\ \cos C = \tfrac{7}{25}\end{gathered}\]

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Exercise 8.1, Question 2

\(\triangle PQR\) is right-angled at Q with \(PQ = 12\) cm and \(PR = 13\) cm. Find \(\tan P - \cot R\).
Show solution
  1. \(QR = \sqrt{169 - 144} = 5\).
  2. \(\tan P = \dfrac{QR}{PQ} = \dfrac{5}{12}\), \(\cot R = \dfrac{QR}{PQ} = \dfrac{5}{12}\).
Answer: \(0\)

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Exercise 8.1, Question 3

If \(\sin A = \tfrac34\), find \(\cos A\) and \(\tan A\).
Show solution
  1. Opposite 3, hypotenuse 4, so adjacent \(= \sqrt{16 - 9} = \sqrt7\).
Answer: \[\begin{gathered}\cos A = \tfrac{\sqrt7}{4}, \\ \tan A = \tfrac{3}{\sqrt7}\end{gathered}\]

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Exercise 8.1, Question 4

If \(15\cot A = 8\), find \(\sin A\) and \(\sec A\).
Show solution
  1. \(\cot A = \tfrac{8}{15}\): adjacent 8, opposite 15, hypotenuse \(\sqrt{64 + 225} = 17\).
Answer: \[\begin{gathered}\sin A = \tfrac{15}{17}, \\ \sec A = \tfrac{17}{8}\end{gathered}\]

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Exercise 8.1, Question 5

If \(\sec\theta = \tfrac{13}{12}\), find the other trigonometric ratios.
Show solution
  1. Adjacent 12, hypotenuse 13, opposite \(\sqrt{169 - 144} = 5\).
Answer: \[\begin{gathered}\cos\theta = \tfrac{12}{13}, \\ \sin\theta = \tfrac{5}{13}, \\ \tan\theta = \tfrac{5}{12}, \\ \cot\theta = \tfrac{12}{5}, \\ \csc\theta = \tfrac{13}{5}\end{gathered}\]

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Exercise 8.1, Question 6

If \(\angle A\) and \(\angle B\) are acute and \(\cos A = \cos B\), show \(\angle A = \angle B\).
Show solution
  1. Take a right triangle ACB with the right angle at C: \(\cos A = \dfrac{AC}{AB}\) and \(\cos B = \dfrac{BC}{AB}\).
  2. \(\cos A = \cos B \Rightarrow AC = BC\), so the triangle is isosceles and the angles opposite the equal sides are equal: \(\angle B = \angle A\).
Answer: Proved.

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Exercise 8.1, Question 7

If \(\cot\theta = \tfrac78\), evaluate:
(i) \(\dfrac{(1 + \sin\theta)(1 - \sin\theta)}{(1 + \cos\theta)(1 - \cos\theta)}\)
Show solution
  1. The fraction is \[\begin{aligned}\dfrac{1 - \sin^2\theta}{1 - \cos^2\theta} &= \dfrac{\cos^2\theta}{\sin^2\theta} \\ &= \cot^2\theta\end{aligned}\]
Answer: \(\tfrac{49}{64}\)
(ii) \(\cot^2\theta\)
Show solution
  1. \(\left(\tfrac78\right)^2\).
Answer: \(\tfrac{49}{64}\)

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Exercise 8.1, Question 8

If \(3\cot A = 4\), check whether \(\dfrac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A\).
Show solution
  1. \(\tan A = \tfrac34\); hypotenuse 5, so \(\sin A = \tfrac35\), \(\cos A = \tfrac45\).
  2. Left: \[\dfrac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \dfrac{7}{25}\] Right: \[\tfrac{16}{25} - \tfrac{9}{25} = \tfrac{7}{25}\]
Answer: Yes, both sides equal \(\tfrac{7}{25}\)

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Exercise 8.1, Question 9

In \(\triangle ABC\), right-angled at B, \(\tan A = \dfrac{1}{\sqrt3}\). Find:
(i) \(\sin A\cos C + \cos A\sin C\)
Show solution
  1. Take \(BC = 1\), \(AB = \sqrt3\), \(AC = 2\): \[\begin{gathered}\sin A = \tfrac12, \\ \cos A = \tfrac{\sqrt3}{2}, \\ \sin C = \tfrac{\sqrt3}{2}, \\ \cos C = \tfrac12\end{gathered}\]
  2. \[\tfrac12 \cdot \tfrac12 + \tfrac{\sqrt3}{2} \cdot \tfrac{\sqrt3}{2} = \tfrac14 + \tfrac34\]
Answer: \(1\)
(ii) \(\cos A\cos C - \sin A\sin C\)
Show solution
  1. \[\tfrac{\sqrt3}{2} \cdot \tfrac12 - \tfrac12 \cdot \tfrac{\sqrt3}{2}\]
Answer: \(0\)

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Exercise 8.1, Question 10

In \(\triangle PQR\), right-angled at Q, \(PR + QR = 25\) cm and \(PQ = 5\) cm. Find \(\sin P\), \(\cos P\) and \(\tan P\).
Show solution
  1. \(PR^2 - QR^2 = PQ^2 = 25\), so \[\begin{aligned}&(PR - QR)(PR + QR) = 25 \\ \Rightarrow\ &PR - QR = 1\end{aligned}\]
  2. \(PR = 13\), \(QR = 12\).
Answer: \[\begin{gathered}\sin P = \tfrac{12}{13}, \\ \cos P = \tfrac{5}{13}, \\ \tan P = \tfrac{12}{5}\end{gathered}\]

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Exercise 8.1, Question 11

True or false? Justify.
(i) \(\tan A\) is always less than 1.
Show solution
  1. \(\tan 60^\circ = \sqrt3 > 1\).
Answer: False
(ii) \(\sec A = \tfrac{12}{5}\) for some angle A.
Show solution
  1. \(\sec A \ge 1\) for acute A, and \(\tfrac{12}{5} > 1\): take hypotenuse 12, adjacent 5.
Answer: True
(iii) \(\cos A\) is the abbreviation of the cosecant of A.
Show solution
  1. \(\cos\) is cosine; cosecant is written \(\csc\) or cosec.
Answer: False
(iv) \(\cot A\) is the product of cot and A.
Show solution
  1. \(\cot A\) is one symbol: the cotangent of angle A.
Answer: False
(v) \(\sin\theta = \tfrac43\) for some angle \(\theta\).
Show solution
  1. \[\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}} \le 1\], and \(\tfrac43 > 1\).
Answer: False

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Introduction to Trigonometry in our sample papers: Sample paper 1 (questions 10, 24, 30) · Sample paper 2 (questions 11, 12, 24, 30) · Sample paper 3 (questions 13, 14, 23, 30) · Sample paper 4 (questions 12, 24, 31) · Sample paper 5 (questions 13, 14, 20, 23, 30).

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