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Q6:
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 in a right-angled triangle, such that $\cos A = \cos B$.
To Prove: $\angle A = \angle B$.
Step 1: Defining the Trigonometric Ratios
Consider a right-angled triangle $ABC$ where $\angle C = 90^\circ$. Since $\angle A$ and $\angle B$ are acute angles, they are the other two angles of the triangle.
By the definition of the cosine ratio in a right-angled triangle:
$\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}$
Step 2: Expressing $\cos A$ and $\cos B$
For $\angle A$:
The side adjacent to $\angle A$ is $AC$.
The hypotenuse is $AB$.
Therefore, $\cos A = \frac{AC}{AB}$ --- (Equation 1)
For $\angle B$:
The side adjacent to $\angle B$ is $BC$.
The hypotenuse is $AB$.
Therefore, $\cos B = \frac{BC}{AB}$ --- (Equation 2)
Step 3: Equating the Ratios
Given that $\cos A = \cos B$, we substitute the expressions from Equation 1 and Equation 2:
$\frac{AC}{AB} = \frac{BC}{AB}$
Step 4: Algebraic Simplification
Multiply both sides of the equation by $AB$ (since $AB$ is a side length of a triangle, $AB \neq 0$):
$AC = BC$
Step 5: Applying Geometric Theorems
In $\triangle ABC$, we have shown that side $AC = BC$.
According to the Isosceles Triangle Theorem, angles opposite to equal sides of a triangle are equal.
Since $AC$ is the side opposite to $\angle B$ and $BC$ is the side opposite to $\angle A$:
$\angle A = \angle B$
Final Answer: Since the sides opposite to the equal angles are equal, it is proven that $\angle A = \angle B$.
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.
- 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
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