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Q9(i):
In triangle ABC, right-angled at B, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of: (i) $\sin A \cos C + \cos A \sin C$
Solution :
Given: In $\triangle ABC$, $\angle B = 90^\circ$ and $\tan A = \frac{1}{\sqrt{3}}$.
To find: The value of $\sin A \cos C + \cos A \sin C$.
Step 1: Determine the sides of the triangle.
In a right-angled triangle, $\tan A = \frac{\text{Opposite side to } A}{\text{Adjacent side to } A} = \frac{BC}{AB}$.
Given $\tan A = \frac{1}{\sqrt{3}}$, we can assume $BC = 1k$ and $AB = \sqrt{3}k$, where $k$ is a positive constant.
Step 2: Calculate the hypotenuse using the Pythagoras Theorem.
The Pythagoras Theorem states: $AC^2 = AB^2 + BC^2$.
$AC^2 = (\sqrt{3}k)^2 + (1k)^2$
$AC^2 = 3k^2 + 1k^2 = 4k^2$
$AC = \sqrt{4k^2} = 2k$.
Step 3: Determine the trigonometric ratios.
For angle $A$:
$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{1k}{2k} = \frac{1}{2}$
$\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}$
For angle $C$:
$\sin C = \frac{\text{Opposite to } C}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}$
$\cos C = \frac{\text{Adjacent to } C}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{1k}{2k} = \frac{1}{2}$
Step 4: Evaluate the expression $\sin A \cos C + \cos A \sin C$.
Substitute the values obtained in Step 3:
$= (\frac{1}{2}) \times (\frac{1}{2}) + (\frac{\sqrt{3}}{2}) \times (\frac{\sqrt{3}}{2})$
$= \frac{1}{4} + \frac{3}{4}$
$= \frac{1 + 3}{4}$
$= \frac{4}{4} = 1$.
Final Answer: 1
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 $\sec A$.
- 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(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
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