Ganit Prakash - Class-X - Trigonometry
Let us work out 23.1 Trigonometric Ratios
📘 Exercise 23.1 Solutions (Q1 - Q9)
Trigonometric RatiosExplanation:
In the right-angled triangle ABC, with respect to the angle $\angle ABC$:
Perpendicular (opposite side) = $AC = 6\text{ cm.}$
Base (adjacent side) = $BC = 8\text{ cm.}$
Hypotenuse = $AB = 10\text{ cm.}$
The sine of $\angle ABC$ is the ratio of the perpendicular to the hypotenuse:
\[\begin{array}{l} \sin(\angle ABC) = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AC}{AB} \\ = \frac{6}{10} = \mathbf{\frac{3}{5}} \end{array}\]The tangent of $\angle ABC$ is the ratio of the perpendicular to the base:
\[\begin{array}{l} \tan(\angle ABC) = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AC}{BC} \\ = \frac{6}{8} = \mathbf{\frac{3}{4}} \end{array}\]Explanation:
In the right-angled triangle ABC, $\angle ABC = 90^\circ$.
With respect to angle $A$:
Base = $AB = 24\text{ cm.}$
Perpendicular = $BC = 7\text{ cm.}$
By Pythagoras theorem, the hypotenuse $AC$ is:
\[\begin{array}{l} AC = \sqrt{AB^2 + BC^2} \\ = \sqrt{24^2 + 7^2} \\ = \sqrt{576 + 49} \\ = \sqrt{625} \\ = 25\text{ cm.} \end{array}\]Now, calculating the required trigonometric ratios for angle $A$:
\[\begin{array}{l} \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} = \mathbf{\frac{7}{25}} \\ \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{AC} = \mathbf{\frac{24}{25}} \\ \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \mathbf{\frac{7}{24}} \\ \operatorname{cosec} A = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{AC}{BC} = \mathbf{\frac{25}{7}} \end{array}\]Explanation:
In the right-angled triangle ABC, $\angle C = 90^\circ$. So, $AB$ is the hypotenuse.
Given: Hypotenuse $AB = 29\text{ units}$, side $BC = 21\text{ units}$.
By Pythagoras theorem, the other side $AC$ is:
\[\begin{array}{l} AC = \sqrt{AB^2 - BC^2} \\ = \sqrt{29^2 - 21^2} \\ = \sqrt{(29 + 21)(29 - 21)} \\ = \sqrt{50 \times 8} \\ = \sqrt{400} \\ = 20\text{ units} \end{array}\]With respect to angle $A$:
Perpendicular = $BC = 21\text{ units}$, Base = $AC = 20\text{ units}$, Hypotenuse = $AB = 29\text{ units}$.
\[\begin{array}{l} \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AB} = \mathbf{\frac{21}{29}} \\ \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AC}{AB} = \mathbf{\frac{20}{29}} \end{array}\]With respect to angle $B$:
Perpendicular = $AC = 20\text{ units}$, Base = $BC = 21\text{ units}$, Hypotenuse = $AB = 29\text{ units}$.
\[\begin{array}{l} \sin B = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AC}{AB} = \mathbf{\frac{20}{29}} \\ \cos B = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{BC}{AB} = \mathbf{\frac{21}{29}} \end{array}\]Explanation:
Given:
\[\begin{array}{l} \cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{7}{25} \end{array}\]Let the Base be $7k$ and the Hypotenuse be $25k$ (where $k > 0$ is a constant).
By Pythagoras theorem, the Perpendicular is:
\[\begin{array}{l} \text{Perpendicular} = \sqrt{\text{Hypotenuse}^2 - \text{Base}^2} \\ = \sqrt{(25k)^2 - (7k)^2} \\ = \sqrt{625k^2 - 49k^2} \\ = \sqrt{576k^2} \\ = 24k \end{array}\]Now, we can find the values of all other trigonometric ratios for the angle $\theta$:
\[\begin{array}{l} \sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{24k}{25k} = \mathbf{\frac{24}{25}} \\ \tan\theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{24k}{7k} = \mathbf{\frac{24}{7}} \\ \operatorname{cosec}\theta = \frac{1}{\sin\theta} = \mathbf{\frac{25}{24}} \\ \sec\theta = \frac{1}{\cos\theta} = \mathbf{\frac{25}{7}} \\ \cot\theta = \frac{1}{\tan\theta} = \mathbf{\frac{7}{24}} \end{array}\]Explanation:
Given $\cot\theta = 2$. We know $\cot\theta = \frac{\text{Base}}{\text{Perpendicular}}$.
Let Base $= 2k$ and Perpendicular $= 1k$ (where $k > 0$).
By Pythagoras theorem, Hypotenuse $= \sqrt{(2k)^2 + (1k)^2} = \sqrt{4k^2 + k^2} = \sqrt{5}k$.
Now, let's find the required values:
\[\begin{array}{l} \tan\theta = \frac{1}{\cot\theta} = \mathbf{\frac{1}{2}} \\ \sec\theta = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{\sqrt{5}k}{2k} = \mathbf{\frac{\sqrt{5}}{2}} \end{array}\]To show: $1 + \tan^2\theta = \sec^2\theta$
\[\begin{array}{l} \text{L.H.S.} = 1 + \tan^2\theta = 1 + \left(\frac{1}{2}\right)^2 = 1 + \frac{1}{4} = \frac{5}{4} \\ \text{R.H.S.} = \sec^2\theta = \left(\frac{\sqrt{5}}{2}\right)^2 = \frac{5}{4} \end{array}\]Since L.H.S. = R.H.S., the relation is proved.
Explanation:
Given $\cos\theta = 0.6 = \frac{6}{10} = \frac{3}{5}$.
We know $\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}}$. Let Base $= 3k$ and Hypotenuse $= 5k$ ($k > 0$).
By Pythagoras theorem, Perpendicular $= \sqrt{(5k)^2 - (3k)^2} = \sqrt{25k^2 - 9k^2} = \sqrt{16k^2} = 4k$.
Now find $\sin\theta$ and $\tan\theta$:
\[\begin{array}{l} \sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{4k}{5k} = \frac{4}{5} \\ \tan\theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{4k}{3k} = \frac{4}{3} \end{array}\]To show: $5\sin\theta - 3\tan\theta = 0$
\[\begin{array}{l} \text{L.H.S.} = 5 \left(\frac{4}{5}\right) - 3 \left(\frac{4}{3}\right) \\ = 4 - 4 \\ = 0 = \text{R.H.S.} \end{array}\]Hence proved.
Explanation:
Given $\cot A = \frac{4}{7.5} = \frac{40}{75} = \frac{8}{15}$.
$\cot A = \frac{\text{Base}}{\text{Perpendicular}}$. Let Base $= 8k$ and Perpendicular $= 15k$ ($k > 0$).
Hypotenuse $= \sqrt{(8k)^2 + (15k)^2} = \sqrt{64k^2 + 225k^2} = \sqrt{289k^2} = 17k$.
Now find the values:
\[\begin{array}{l} \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{8k}{17k} = \mathbf{\frac{8}{17}} \\ \operatorname{cosec} A = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{17k}{15k} = \mathbf{\frac{17}{15}} \end{array}\]To show: $1 + \cot^2 A = \operatorname{cosec}^2 A$
\[\begin{array}{l} \text{L.H.S.} = 1 + \left(\frac{8}{15}\right)^2 = 1 + \frac{64}{225} = \frac{225 + 64}{225} = \frac{289}{225} \\ \text{R.H.S.} = \left(\frac{17}{15}\right)^2 = \frac{289}{225} \end{array}\]Since L.H.S. = R.H.S., the relation is proved.
Explanation:
Given $\sin C = \frac{2}{3}$. We know $\sin C = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$.
Let Perpendicular $= 2k$ and Hypotenuse $= 3k$ ($k > 0$).
Base $= \sqrt{(3k)^2 - (2k)^2} = \sqrt{9k^2 - 4k^2} = \sqrt{5}k$.
Now, calculate $\cos C$ and $\operatorname{cosec} C$:
\[\begin{array}{l} \cos C = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{\sqrt{5}k}{3k} = \frac{\sqrt{5}}{3} \\ \operatorname{cosec} C = \frac{1}{\sin C} = \frac{3}{2} \end{array}\]Finding the value of the expression:
\[\begin{array}{l} \cos C \times \operatorname{cosec} C = \frac{\sqrt{5}}{3} \times \frac{3}{2} = \mathbf{\frac{\sqrt{5}}{2}} \end{array}\](i) The value of $\tan A$ is always greater than 1.
(ii) The value of $\cot A$ is always less than 1.
(iii) For an angle $\theta$, it may be possible that, $\sin\theta = \frac{4}{3}$.
(iv) For an angle $\alpha$, it may be possible that, $\sec\alpha = \frac{12}{5}$.
(v) For an angle $\beta$ (Beta), it may be possible that, $\operatorname{cosec}\beta = \frac{5}{13}$.
(vi) For an angle $\theta$, it may be possible that, $\cos\theta = \frac{3}{5}$.
Explanations:
(i) False. $\tan A = \frac{\text{Perpendicular}}{\text{Base}}$. In a right triangle, the perpendicular can be greater than, less than, or equal to the base. Therefore, $\tan A$ can take any positive value (e.g., if Perpendicular $= 3$ and Base $= 4$, $\tan A = 0.75$).
(ii) False. $\cot A = \frac{\text{Base}}{\text{Perpendicular}}$. Similar to tangent, the base can be greater than the perpendicular. If Base $= 4$ and Perpendicular $= 3$, then $\cot A = \frac{4}{3} > 1$. So it is not always less than 1.
(iii) False. $\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$. In any right-angled triangle, the hypotenuse is always the longest side. Therefore, the ratio of perpendicular to hypotenuse must be less than 1. Here, $\frac{4}{3} > 1$, which is impossible.
(iv) True. $\sec\alpha = \frac{\text{Hypotenuse}}{\text{Base}}$. Since the hypotenuse is the longest side, this ratio is always greater than 1. Here, $\frac{12}{5} > 1$, which is possible.
(v) False. $\operatorname{cosec}\beta = \frac{\text{Hypotenuse}}{\text{Perpendicular}}$. Since the hypotenuse must be greater than the perpendicular, this ratio must be greater than 1. Here, $\frac{5}{13} < 1$, which is impossible.
(vi) True. $\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}}$. Since the base is less than the hypotenuse, this ratio must be less than 1. Here, $\frac{3}{5} < 1$, which is a valid possible value.
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