Class 10 Mathematics – Chapter 8: Introduction to Trigonometry

Exercise 8.1 – NCERT Solutions

Q1. In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine (i) sin A, cos A (ii) sin C, cos C.

By Pythagoras theorem,

\(AC=\sqrt{24^2+7^2}=\sqrt{625}=25\text{ cm}\).

For angle A, opposite = BC = 7, adjacent = AB = 24 and hypotenuse = AC = 25.

\(\sin A=\frac{7}{25},\quad \cos A=\frac{24}{25}\).

For angle C, opposite = AB = 24 and adjacent = BC = 7.

\(\sin C=\frac{24}{25},\quad \cos C=\frac{7}{25}\).

Q2. In Fig. 8.13, find tan P − cot R.

From Fig. 8.13, \(PQ=12\) cm, \(PR=13\) cm and the right angle is at Q. Hence, by Pythagoras theorem, \(QR=5\) cm.

For angle P, \(\tan P=\frac{QR}{PQ}=\frac{5}{12}\).

For angle R, \(\cot R=\frac{QR}{PQ}=\frac{5}{12}\).

Therefore,

\(\tan P-\cot R=\frac{5}{12}-\frac{5}{12}=0\).

Answer: 0

Q3. If sin A = 3/4, calculate cos A and tan A.

Using \(\sin^2A+\cos^2A=1\),

\(\cos A=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt7}{4}\).

Therefore,

\(\tan A=\frac{\sin A}{\cos A}=\frac{3/4}{\sqrt7/4}=\frac{3}{\sqrt7}=\frac{3\sqrt7}{7}\).

Answer: \(\cos A=\frac{\sqrt7}{4},\;\tan A=\frac{3\sqrt7}{7}\)

Q4. Given 15 cot A = 8, find sin A and sec A.

\(\cot A=\frac{8}{15}\), so take adjacent side = 8k and opposite side = 15k.

The hypotenuse is \(\sqrt{8^2+15^2}k=17k\).

Hence, \(\sin A=\frac{15}{17}\) and \(\sec A=\frac{17}{8}\).

Q5. Given sec θ = 13/12, calculate all other trigonometric ratios.

Since \(\sec\theta=\frac{13}{12}\), take hypotenuse = 13k and adjacent side = 12k.

Opposite side = \(\sqrt{13^2-12^2}k=5k\).

Therefore,

\(\sin\theta=\frac5{13},\;\cos\theta=\frac{12}{13},\;\tan\theta=\frac5{12}\).

Also, \(\cot\theta=\frac{12}{5},\;\cosec\theta=\frac{13}{5}\).

Q6. If ∠A and ∠B are acute angles such that cos A = cos B, show that ∠A = ∠B.

Let two right triangles have acute angles A and B and suppose \(\cos A=\cos B\).

Thus, the ratio of adjacent side to hypotenuse is the same for both angles. By Pythagoras theorem, the corresponding opposite-to-hypotenuse ratios are also equal. Hence the two right triangles are similar by SSS/SAS ratio correspondence, giving equal corresponding acute angles.

Therefore, \(\boxed{\angle A=\angle B}\).

Q7. If cot θ = 7/8, evaluate (i) the given fraction (ii) cot²θ.

(i)

\(\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)} =\frac{1-\sin^2\theta}{1-\cos^2\theta} =\frac{\cos^2\theta}{\sin^2\theta} =\cot^2\theta\).

Since \(\cot\theta=\frac78\), the value is \(\frac{49}{64}\).

(ii) \(\cot^2\theta=\left(\frac78\right)^2=\frac{49}{64}\).

Q8. If 3 cot A = 4, check whether the given expression equals cos²A − sin²A.

From \(3\cot A=4\), we get \(\cot A=\frac43\), so \(\tan A=\frac34\).

The left-hand side is

\(\frac{1-\tan^2A}{1+\tan^2A} =\frac{1-9/16}{1+9/16} =\frac{7/16}{25/16}=\frac7{25}\).

Taking a right triangle with opposite : adjacent = 3 : 4, hypotenuse = 5. Thus \(\sin A=\frac35\), \(\cos A=\frac45\).

Hence \(\cos^2A-\sin^2A=\frac{16}{25}-\frac9{25}=\frac7{25}\).

Both sides are equal. Therefore, the statement is true.

Q9. In triangle ABC, right-angled at B, if tan A = 1/√3, find (i) sin A cos C + cos A sin C (ii) cos A cos C − sin A sin C.

Since \(A+C=90^\circ\), we use the standard identities.

(i) \(\sin A\cos C+\cos A\sin C=\sin(A+C)=\sin90^\circ=1\).

(ii) \(\cos A\cos C-\sin A\sin C=\cos(A+C)=\cos90^\circ=0\).

Q10. In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine sin P, cos P and tan P.

Let \(QR=x\) cm. Then \(PR=25-x\) cm.

By Pythagoras theorem,

\((25-x)^2=5^2+x^2\).

\(625-50x+x^2=25+x^2\), so \(x=12\).

Thus, \(QR=12\) cm and \(PR=13\) cm.

Therefore, \(\sin P=\frac{12}{13},\quad\cos P=\frac5{13},\quad\tan P=\frac{12}{5}\).

Q11. State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1. False. For example, \(\tan60^\circ=\sqrt3>1\).

(ii) \(\sec A=\frac{12}{5}\) for some value of A. True. Since \(\cos A=\frac5{12}\) is possible for an acute angle.

(iii) cos A is the abbreviation for cosecant A. False. cos A denotes cosine, while cosecant is written as cosec A.

(iv) cot A is the product of cot and A. False. cot A denotes the cotangent of angle A.

(v) \(\sin\theta=\frac43\) for some angle θ. False. For a real angle, the value of sine cannot exceed 1.