Class X Session 2023-24 Question 24

 MATH SAMPLE QUESTION PAPER

Class X Session 2023-24

MATHEMATICS STANDARD (Code No.041) 

Question 24

\[if\; \tan(A + B) =\;\sqrt{3} \; and \;\tan(A - B) =\;\frac{1}{\sqrt{3}}\] ,0° < A + B < 90°; A > B, find A and B.

Explanation:  

Given that :

\[ \tan(A + B) =\;\sqrt{3} \; and \;\tan(A - B) =\;\frac{1}{\sqrt{3}}\]

First, recognize the angles for which

\[\tan(\theta) = \sqrt{3} \; and \; tan(\theta) = \frac{1}{\sqrt{3}}\] we know that

\[\tan(60^{\circ}) = \sqrt{3} \; and \; tan(30^{\circ}) = \frac{1}{\sqrt{3}}\]

\[\ so,\; A+B \;= 60^{\circ} \; and \; A-B \;= 30^{\circ}\] now add the two equation , we get 

\[\ A+B \;= 60^{\circ},---(1)\] 

\[\ A-B \;= 30^{\circ},---(2)\]

\[\ so,\;(A+B)+(A-B)\;= 60^{\circ} + 30^{\circ}\]

\[\ or,\;2A\;= 90^{\circ}\]

\[\ or,\;A\;= 45^{\circ}\]

Now Put A value in equation (1) and we get 

\[\ from (1),\;B\;= 60^{\circ} - 45^{\circ}\]

\[\therefore \; B\;= 15^{\circ}\]

\[Now \; A\;= 45^{\circ}\; and \; B\;= 15^{\circ}\]