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Q3:
If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$.
Solution :
Given: In a right-angled triangle $ABC$, where $\angle B = 90^\circ$, we are given that $\sin A = \frac{3}{4}$.
To find: The values of $\cos A$ and $\tan A$.
Step 1: Understanding the Trigonometric Ratio
By definition, in a right-angled triangle, the sine of an angle is the ratio of the side opposite to the angle to the hypotenuse.
$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC}$.
Given $\sin A = \frac{3}{4}$, we can assume $BC = 3k$ and $AC = 4k$, where $k$ is a positive constant.
Step 2: Applying the Pythagoras Theorem
In $\triangle ABC$, by the Pythagoras Theorem:
$AC^2 = AB^2 + BC^2$
Substituting the known values:
$(4k)^2 = AB^2 + (3k)^2$
$16k^2 = AB^2 + 9k^2$
$AB^2 = 16k^2 - 9k^2$
$AB^2 = 7k^2$
$AB = \sqrt{7k^2} = k\sqrt{7}$
Step 3: Calculating $\cos A$
By definition, $\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC}$.
Substituting the values:
$\cos A = \frac{k\sqrt{7}}{4k}$
$\cos A = \frac{\sqrt{7}}{4}$
Step 4: Calculating $\tan A$
By definition, $\tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB}$.
Substituting the values:
$\tan A = \frac{3k}{k\sqrt{7}}$
$\tan A = \frac{3}{\sqrt{7}}$
To rationalize the denominator:
$\tan A = \frac{3}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{3\sqrt{7}}{7}$
Final Answer: $\cos A = \frac{\sqrt{7}}{4}$ and $\tan A = \frac{3\sqrt{7}}{7}$
More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.1
- Q1(i): In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Determine : (i) $\sin A, \cos A$
- Q1(ii): In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Determine : (ii) $\sin C, \cos C$
- Q10: In $\triangle PQR$, right-angled at $Q$, $PR + QR = 25$ cm and $PQ = 5$ cm. Determine the values of $\sin P, \cos P$ and $\tan P$.
- Q11(i): State whether the following are true or false. Justify your answer. (i) The value of $\tan A$ is always less than 1.
- Q11(ii): State whether the following are true or false. Justify your answer. (ii) $\sec A = \frac{12}{5}$ for some value of angle $A$.
- Q11(iii): State whether the following are true or false. Justify your answer. (iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$.
- Q11(iv): State whether the following are true or false. Justify your answer. (iv) $\cot A$ is the product of cot and $A$.
- Q11(v): State whether the following are true or false. Justify your answer. (v) $\sin \theta = \frac{4}{3}$ for some angle $\theta$.
- Q2: In Fig. 8.13, find $\tan P – \cot R$.
- Q4: Given $15 \cot A = 8$, find $\sin A$ and $\sec A$.
- Q5: Given $\sec \theta = \frac{13}{12}$, calculate all other trigonometric ratios.
- Q6: If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$.
- Q7(i): If $\cot \theta = \frac{7}{8}$, evaluate : (i) $\frac{(1 + \sin \theta) (1 - \sin \theta)}{(1 + \cos \theta) (1 - \cos \theta)}$
- Q7(ii): If $\cot \theta = \frac{7}{8}$, evaluate : (ii) $\cot^2 \theta$
- Q8: If $3 \cot A = 4$, check whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A – \sin^2 A$ or not.
- Q9(i): In triangle ABC, right-angled at B, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of: (i) $\sin A \cos C + \cos A \sin C$
- Q9(ii): In triangle ABC, right-angled at B, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of: (ii) $\cos A \cos C – \sin A \sin C$
CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry
Chapters in CBSE - Class 10 Mathematics
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