Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

In a $\Delta ABC$ , the sides a=4,b=2\ and\ \angle $c=90^{\circ}$ . To find the value of $A-B$ .

Option: 1
$36.8^{\circ}$

Option: 2
$35.8^{\circ}$

Option: 3
$37.8^{\circ}$

Option: 4
$34.8^{\circ}$

Answers (1)

best_answer

Answer: Option (a) $36.8^{\circ}$

Explanation:

Given that,

$a=4,b=2\ \text{and}\ \angle c=90^{\circ}$

We know that,

$\angle A+\angle B+\angle C=180^{\circ}$

$\angle A+\angle B=180^{\circ}-\angle C$

$\angle A+\angle B=90^{\circ}$
By the law of tangents,

$\frac{a-b}{a+b}=\frac{tan\left ( \frac{A-B}{2} \right )}{tan\left ( \frac{A+B}{2} \right )}$

$\frac{4-2}{4+2}=\frac{tan\ \frac{1}{2}\left ( A-B \right )}{tan\ \frac{1}{2}\left ( 90^{\circ} \right )}$

$\frac{1}{3}=\frac{tan\ \frac{1}{2}\left ( A-B \right )}{tan\ \left ( 45^{\circ} \right )}$

$\frac{1}{3}tan\ \left ( 45^{\circ} \right )=tan\ \frac{1}{2}\left ( A-B \right )$

$tan\ \frac{1}{2}\left ( A-B \right )=0.3333$

$\frac{1}{2}\left ( A-B \right )=18.4^{\circ}$

$\left ( A-B \right )=36.8^{\circ}$

Posted by

Suraj Bhandari

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

JEE Main 2026: NTA directs non-Aadhaar applicants to submit photo verification certificate by January 7

anam-khan

Jharkhand CM inaugurates free coaching institute for tribal medical, engineering students

anam-khan

JEE Main 2026: IIT Roorkee’s ECE emerges as top choice after CSE; closing rank rises

Latest Question