Find the best tutors and institutes for Class 10 Tuition
Q5:
Given $\sec \theta = \frac{13}{12}$, calculate all other trigonometric ratios.
Solution :
Given: $\sec \theta = \frac{13}{12}$
To find: All other trigonometric ratios, i.e., $\sin \theta, \cos \theta, \tan \theta, \csc \theta,$ and $\cot \theta$.
Step 1: Defining the Trigonometric Ratio
In a right-angled triangle, $\sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent side}}$.
Given $\sec \theta = \frac{13}{12}$, let the Hypotenuse ($AC$) = $13k$ and the Adjacent side ($AB$) = $12k$, where $k$ is a positive constant.
Step 2: Finding the Third Side using Pythagoras Theorem
According to the Pythagoras Theorem: $AC^2 = AB^2 + BC^2$
Substituting the known values:
$(13k)^2 = (12k)^2 + BC^2$
$169k^2 = 144k^2 + BC^2$
$BC^2 = 169k^2 - 144k^2$
$BC^2 = 25k^2$
$BC = \sqrt{25k^2} = 5k$ (Perpendicular/Opposite side)
Step 3: Calculating the Trigonometric Ratios
Using the definitions of trigonometric ratios based on the sides of the triangle:
1. $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5k}{13k} = \frac{5}{13}$
2. $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13}$
3. $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12}$
4. $\csc \theta = \frac{1}{\sin \theta} = \frac{13}{5}$
5. $\cot \theta = \frac{1}{\tan \theta} = \frac{12}{5}$
Final Answer: The trigonometric ratios are $\sin \theta = \frac{5}{13}, \cos \theta = \frac{12}{13}, \tan \theta = \frac{5}{12}, \csc \theta = \frac{13}{5},$ and $\cot \theta = \frac{12}{5}$.
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 $\tan P – \cot R$.
- Q3: If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$.
- Q4: Given $15 \cot A = 8$, find $\sin A$ and $\sec A$.
- 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
Download free CBSE - Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.1 worksheets
Download Now
