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Q3:
If $\tan (A + B) = \sqrt{3}$ and $\tan (A – B) = \frac{1}{\sqrt{3}}$; $0^\circ < A + B \le 90^\circ$; $A > B$, find $A$ and $B$.
Solution :
Given:
1. $\tan(A + B) = \sqrt{3}$
2. $\tan(A - B) = \frac{1}{\sqrt{3}}$
3. $0^\circ < A + B \le 90^\circ$
4. $A > B$
To find:
The values of angles $A$ and $B$.
Step 1: Determine the angles using trigonometric values.
We know from the standard trigonometric table that $\tan(60^\circ) = \sqrt{3}$.
Given $\tan(A + B) = \sqrt{3}$, we can equate the angles:
$A + B = 60^\circ$ --- (Equation 1)
Similarly, we know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.
Given $\tan(A - B) = \frac{1}{\sqrt{3}}$, we can equate the angles:
$A - B = 30^\circ$ --- (Equation 2)
Step 2: Solve the system of linear equations.
We have a system of two linear equations in two variables:
(1) $A + B = 60^\circ$
(2) $A - B = 30^\circ$
To solve for $A$, we add Equation 1 and Equation 2:
$(A + B) + (A - B) = 60^\circ + 30^\circ$
$A + A + B - B = 90^\circ$
$2A = 90^\circ$
$A = \frac{90^\circ}{2}$
$A = 45^\circ$
Step 3: Substitute the value of $A$ to find $B$.
Substitute $A = 45^\circ$ into Equation 1:
$45^\circ + B = 60^\circ$
$B = 60^\circ - 45^\circ$
$B = 15^\circ$
Step 4: Verification of constraints.
Check if $A > B$: $45^\circ > 15^\circ$ (True).
Check if $0^\circ < A + B \le 90^\circ$: $45^\circ + 15^\circ = 60^\circ$, and $0^\circ < 60^\circ \le 90^\circ$ (True).
Final Answer: A = 45^\circ, B = 15^\circ
More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.2
- Q1(i): Evaluate the following : (i) $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$
- Q1(ii): Evaluate the following : (ii) $2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$
- Q1(iii): Evaluate the following : (iii) $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$
- Q1(iv): Evaluate the following : (iv) $\frac{\sin 30^\circ + \tan 45^\circ – \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$
- Q1(v): Evaluate the following : (v) $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$
- Q2(i): Choose the correct option and justify your choice : (i) $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} =$
- Q2(ii): Choose the correct option and justify your choice : (ii) $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} =$
- Q2(iii): Choose the correct option and justify your choice : (iii) $\sin 2A = 2 \sin A$ is true when $A =$
- Q2(iv): Choose the correct option and justify your choice : (iv) $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} =$
- Q4(i): State whether the following are true or false. Justify your answer. (i) $\sin (A + B) = \sin A + \sin B$.
- Q4(ii): State whether the following are true or false. Justify your answer. (ii) The value of $\sin \theta$ increases as $\theta$ increases.
- Q4(iii): State whether the following are true or false. Justify your answer. (iii) The value of $\cos \theta$ increases as $\theta$ increases.
- Q4(iv): State whether the following are true or false. Justify your answer. (iv) $\sin \theta = \cos \theta$ for all values of $\theta$.
- Q4(v): State whether the following are true or false. Justify your answer. (v) $\cot A$ is not defined for $A = 0^\circ$.
CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry
Chapters in CBSE - Class 10 Mathematics
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