EXERCISE 8.1
1. In triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C
Solution: Given,
In triangle ABC, angle B is a right angle. The lengths of two sides are given: AB=24 cm, BC = 7 cm, and AC=?
Now by Using the Pythagorean Theorem, we have
\(AC^2 = AB^2 + BC^2\)
\( AC^2 = 24^2 + 7^2 = 576 + 49 = 625 \)
\( AC = \sqrt{625} = 25\,\text{cm} \)
Now we know all the sides:
AB = 24 cm
BC = 7 cm
AC = 25 cm
Therefore, (i) sin A, cos A
For \(\theta\) is in A, Perpendicular = BC and Base = AB, as right-angled at B.
\( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \)
= \( \frac{BC}{AC} \) = \( \frac{7cm}{25 cm} \) = = \( \frac{7}{25} \)
and \( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} \)
= \( \frac{AB}{AC} \) = \( \frac{24cm}{25 cm} \)
= \( \frac{24}{25} \)
Now for (ii) sin C, cos C
For \(\theta\) is in C, Perpendicular = AB and Base = BC as right-angled at B.
\( \sin C = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \)
= \( \frac{AB}{AC} \) = \( \frac{24cm}{25 cm} \) = \( \frac{24}{25} \)
and \( \cos C = \frac{\text{Base}}{\text{Hypotenuse}} \)
= \( \frac{BC}{AC} \) = \( \frac{7cm}{25 cm} \) = \( \frac{7}{25} \)
2. In Fig. 8.13, find tan P – cot R
Solution: In Fig 8.13, given PQ = 12 cm and PR = 13 cm, QR = ?, and right-angle at Q.
Now by using the Pythagorean Theorem, we have:
\[ PR^2 = PQ^2 + QR^2 \]
\[ 13^2 = 12^2 + QR^2 \]
\[169 = 144 + QR^2 \]
\[QR^2 = 169 – 144 = 25\]
\[QR = \sqrt{25} = 5\ \text{cm}\]
\[\tan P = \frac{\text{adjacent to } P}{\text{opposite to } P} \]
\[ = \frac{QR}{PQ} = \frac{5}{12} \]
\[\cot R = \frac{\text{opposite to } R}{\text{adjacent to } R} \]
\[ = \frac{QR}{PQ} = \frac{5}{12} \]
Therefore, tan P – cot R = \[\tan P – \cot R = \frac{5}{12} – \frac{5}{12} = 0\]
3. If sin A = \( \frac{3}{4}\) calculate cos A and tan A
Solution: Given, sin A = \( \frac{3}{4}\)
By definition: \( \sin A = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{3}{4} \)
By using the Pythagorean Theorem, we have
\( \text{base}^2 = \text{hypotenuse}^2 – \text{perpendicular}^2 = 4^2 – 3^2 \)
= 16 – 9 = 7
\( \text{base} = \sqrt{7} \)
Therefore, \[\cos A = \frac{\text{base}}{\text{hypotenuse}} = \frac{\sqrt{7}}{4} \]
and \[\tan A = \frac{\text{perpendicular}}{\text{base}} = \frac{3}{\sqrt{7}} \]
4. Given 15 cot A = 8, find sin A and sec A
Solution: Given 15 cot A = 8
\( \cot A = \frac{8}{15} \)
We know that \( \cot A = \frac{\text{base}}{\text{perpendicular}} \).
\( \cot A = \frac{\text{base}}{\text{perpendicular}} = \frac{8}{15} \)
\[ \text{base} = 8, \quad \text{perpendicular} = 15 \]
Now, using the Pythagorean theorem to find the hypotenuse:
\[ \text{hypotenuse} = \sqrt{8^2 + 15^2} \]
\[ = \sqrt{64 + 225} = \sqrt{289} = 17 \]
Therefore, \( \sin A = \frac{\text{perpendicular}}{\text{hypotenuse}} \)
\( = \frac{15}{17} \)
and \( \sec A = \frac{1}{\cos A} = \frac{\text{hypotenuse}}{\text{base}} \)
\( = \frac{17}{8} \)
5. Given \( sec \theta = \frac{13}{12} \), Calculate all other trigonometric ratios.
Solution: Given \( sec \theta = \frac{13}{12} \)
\( \sec \theta = \frac{\text{hypotenuse}}{\text{base}} = \frac{13}{12} \)
So, \( \frac{\text{hypotenuse}}{\text{base}} = \frac{13}{12} \)
\[ \text{hypotenuse} = 13, \quad \text{base} = 12 \]
Now, by using the Pythagorean theorem we have
\( \text{perpendicular} = \sqrt{13^2 – 12^2} \)
\( = \sqrt{169 – 144} = \sqrt{25} = 5 \)
Therefore, \( \sin \theta = \frac{\text{perpendicular}}{\text{hypotenuse}} \)
\( = \frac{5}{13} \)
\( \cos \theta = \frac{\text{base}}{\text{hypotenuse}} = \frac{12}{13} \)
\( \tan \theta = \frac{\text{perpendicular}}{\text{base}} = \frac{5}{12} \)
\( \cot \theta = \frac{\text{base}}{\text{perpendicular}} = \frac{12}{5} \)
\( \sec \theta = \frac{13}{12} \) (Given)
\( \csc \theta = \frac{\text{hypotenuse}}{\text{perpendicular}} = \frac{13}{5} \)
6. If \( \angle A \) and \( \angle B \) are acute angles such that \( \cos A \) = \( \cos B \), then show that \( \angle A \) = \( \angle B \).
Solution: Given, \(\angle A\) and \(\angle B\) are acute angles such that \(\cos A = \cos B\). \(\textbf{To prove that:} \quad \angle A = \angle B \)
\(\textbf{Proof:} \) Since \(\angle A\) and \(\angle B\) are acute angles, we have \( 0 < A, B < \frac{\pi}{2}\)
The cosine function is strictly decreasing and one-to-one on the interval \(\left(0, \frac{\pi}{2}\right)\). Therefore, if \(\cos A = \cos B\), then it must be that \(\angle A\) = \(\angle B\)
So, \(\angle A\) = \(\angle B\)
Hence proved.
7. If \(\cot \theta = \frac{7}{8}\), evaluate : (i) \( \frac{(1 + \sin \theta)(1 – \sin \theta)}{(1 + \cos \theta)(1 – \cos \theta)} \) and (ii) \( \cot^2 \theta \)
Solution: Given \( \cot \theta = \frac{7}{8} \)
\( \implies \cot \theta = \frac{b}{p} = \frac{7}{8} \)
\( \implies b = 7, \quad p = 8 \)
Now, by using the Pythagorean theorem we have
\( h^2 = p^2 + b^2 \)
\( \implies h^2 = 8^2 + 7^2 \)
\( \implies h = \sqrt{7^2 + 8^2} = \sqrt{49 + 64} = \sqrt{113} \)
So, \( \sin \theta = \frac{p}{h} = \frac{8}{\sqrt{113}} \)
and \( \cos \theta = \frac{b}{h} = \frac{7}{\sqrt{113}} \)
Now, (i) \( \frac{(1 + \sin \theta)(1 – \sin \theta)}{(1 + \cos \theta)(1 – \cos \theta)} \)
\( \implies \frac{(1 + \frac{8}{\sqrt{113}})(1 – \frac{8}{\sqrt{113}})}{(1 + \frac{7}{\sqrt{113}})(1 – \frac{7}{\sqrt{113}})} \)
By using the identity \( (1 + a)(1 – a) = 1 – a^2 \)
\( \implies \frac{1 – \left( \frac{8}{\sqrt{113}} \right)^2}{1 – \left( \frac{7}{\sqrt{113}} \right)^2} \)
\( \implies \frac{1 – \frac{64}{113}}{1 – \frac{49}{113}} \)
\( \implies \frac{\frac{113-64}{113}}{\frac{113-49}{113}} \)
\( \implies \frac{\frac{49}{113}}{\frac{64}{113}} \)
\( \implies \frac{49}{64} \)
and (ii) \( \cot^2 \theta \)
\( \implies \cot^2 \theta = \left( \frac{b}{p} \right)^2 \)
\( = \left( \frac{7}{8} \right)^2 = \frac{49}{64} \)
8. If \(3 \cot A = 4 \), check whether \( \frac{2 \tan A}{1 + \tan^2 A} = \cos^2 A – \sin^2 A \)
Solution: Given, \(3 \cot A = 4 \) = \( \cot A = \frac{4}{3} \)
\( \cot A = \frac{b}{p} = \frac{4}{3} \)
Here, Base = 4 and perpendicular = 3
Now using Pythagoras theorem, we have
\(h^2 = p^2 + b^2 \)
\( \text{Hypotenuse} = \sqrt{3^2 + 4^2} \)
\( = \sqrt{9 + 16} = \sqrt{25} = 5 \)
Now, \( \tan A = \frac{p}{b} = \frac{3}{4} \) and \( \sin A = \frac{p}{h} = \frac{3}{5} \) and \( \cos A = \frac{b}{h} = \frac{4}{5} \)
\( \textbf{LHS:} \frac{2 \tan A}{1 + \tan^2 A} \)
By substituting \( \tan A = \frac{3}{4} \), we have:
\( = \frac{2 \cdot \frac{3}{4}}{1 + \left( \frac{3}{4} \right)^2} \)
\( = \frac{\frac{6}{4}}{1 + \frac{9}{16}} \)
\( = \frac{\frac{3}{2}}{\frac{25}{16}} \)
\( = \frac{3}{2} \cdot \frac{16}{25} \)
\( = \frac{48}{50} \)
\( = \frac{24}{25} \)
\( \textbf{RHS:} \cos^2 A – \sin^2 A \) \( = \left( \frac{4}{5} \right)^2 – \left( \frac{3}{5} \right)^2 \)
\( = \frac{16}{25} – \frac{9}{25} \)
\( = \frac{7}{25} \)
So, \( \frac{2 \tan A}{1 + \tan^2 A} \ne \cos^2 A – \sin^2 A \)
9. In triangle \(ABC\), right-angled at (B), if \(\tan A = \frac{1}{3}\)
\((i) \sin A \cos C + \cos A \sin C\)
\((ii)\cos A \cos C – \sin A \sin C\)
Solution: Given, In triangle \(ABC\), right-angled at (B), and \(\tan A = \frac{1}{3}\)
\(\tan A =\frac{p}{b} = \frac{1}{3}\)
Here, perpendicular = 1 and Base = 3
Now using Pythagoras theorem, we have
\(h^2 = p^2 + b^2 \)
\( \text{Hypotenuse} = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \)
Now \( \sin A = \frac{1}{\sqrt{10}}, \sin C = \frac{3}{\sqrt{10}}\),
\( \cos A = \frac{3}{\sqrt{10}}, \cos C = \frac{1}{\sqrt{10}} \)
Now, \((i) \sin A \cos C + \cos A \sin C\)
\[ \sin A \cos C + \cos A \sin C = \sin A \sin A + \cos A \cos A \]
\[ = \sin^2 A + \cos^2 A = 1 \]
and \((ii)\cos A \cos C – \sin A \sin C\)
\[ \cos A \cos C – \sin A \sin C = \cos A \sin A – \sin A \cos A = 0 \]
10. In triangle \(PQR\), right-angled at \(Q\), \(PR + QR = 25 \, \text{cm}\), and \(PQ = 5 \, \text{cm}\). Determine the values of \(\sin P\), \(\cos P\), and \(\tan P\).
Solution: Given, In triangle \(PQR\), right-angled at \(Q\), \(PR + QR = 25 \text{cm}\), and \(PQ = 5 \, \text{cm}\)
Let \(QR = x\), then \(PR = 25 – x\). Given that \(\triangle PQR\) is right-angled at \(Q\), Now by applying the Pythagorean theorem, we have
\( PR^2 = PQ^2 + QR^2 \)
\( (25 – x)^2 = 5^2 + x^2 \)
\( 625 – 50x + x^2 = 25 + x^2 \) \( 625 – 50x = 25 \)
\( 600 = 50x \Rightarrow x = 12 \)
Therefore \( QR = 12 \, \text{cm}, \quad PR = 25 – 12 = 13 \, \text{cm}, \quad PQ = 5 \, \text{cm} \)
So, \( \sin P = \frac{\text{P}}{\text{H}} = \frac{QR}{PR} = \frac{12}{13} \)
\( \cos P = \frac{\text{B}}{\text{H}} = \frac{PQ}{PR} = \frac{5}{13} \)
\( \tan P = \frac{P}{B} = \frac{QR}{PQ} = \frac{12}{5} \)
11. State whether the following are true or false. Justify your answer
(i) The value of tan A is always less than 1.
Answer: False
Justification: The value of \(\tan A = \frac{\sin A}{\cos A} \) depends on the angle \(A\). In a right-angled triangle, angle \(A\) lies between \(0^\circ\) and \(90^\circ\). As \(A\) increases, \(\tan A\) also increases. When \( \tan 45^\circ = 1 \), If \(A < 45^\circ\), then \(\tan A < 1\). If \(A > 45^\circ\), then \(\tan A > 1\). Therefore, the statement “the value of \(\tan A\) is always less than 1” is \(\textbf{false}\).
(ii) \(\sec A = \frac{12}{5} \text{ for some value of angle } A\)
Answer: True
Justification: The value of \(\sec A = \frac{1}{\cos A}\), and in a right-angled triangle, \(\cos A = \frac{\text{adjacent}}{\text{hypotenuse}}\). If \(\sec A = \frac{12}{5}\), then \(\cos A = \frac{5}{12}\), which is a valid value between 0 and 1. This corresponds to a right-angled triangle where the adjacent side is 5 and the hypotenuse is 12, which is possible since the hypotenuse is the longest side. Therefore, the statement “\(\sec A = \frac{12}{5}\) for some value of angle \(A\)” is \(\textbf{true}\).
(iii) \(\cos A\) is the abbreviation used for the cosecant of angle \(A\)
Answer: False
Justification: The symbol \(\cos A\) stands for the cosine of angle \(A\), not the cosecant. The abbreviation for cosecant is \(\csc A\), which is defined as: \(\csc A = \frac{1}{\sin A}\) Whereas: \(\cos A = \frac{\text{adjacent side}}{\text{hypotenuse}}\). Therefore, the statement “\(\cos A\) is the abbreviation used for the cosecant of angle \(A\)” is \(\textbf{false}\).
(iv) \(cot A\) is the product of \(\cot\) and \(A\)
Answer: False
Justification: \(\cot A\) is not the product of \(\cot\) and \(A\); it is a trigonometric function that represents the cotangent of angle \(A\). \(\cot A\) is defined as: \(\cot A = \frac{1}{\tan A} = \frac{\cos A}{\sin A}\). It is a single function applied to the angle \(A\), not a multiplication. Therefore, the statement “\(\cot A\) is the product of cot and \(A\)” is \(\textbf{false}\).
(v) \(\sin \theta = \frac{4}{3} \text{ for some angle } \theta \)
Answer: False
Justification: The sine of an angle, \(\sin \theta\), represents the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle.. Since the hypotenuse is always the longest side, the value of \(\sin \theta\) must be between -1 and 1: \(-1 \leq \sin \theta \leq 1 \). Given \(\sin \theta = \frac{4}{3} \approx 1.33\), which is greater than 1, this is not possible for any angle \(\theta\). Therefore, the statement “\(\sin \theta = \frac{4}{3}\) for some angle \(\theta\)” is \(\textbf{false}\).
