In a ∆ABC, if ∠C=90°, (a^2+b^2)/(a^2-b^2)sin(A-B)=1 and 0<B<45°, then

Correct answer: (B) c>a>b. We have (a^2+b^2)/(a^2-b^2)sin(A-B)=1 ⇒ sin(A-B)=(a^2-b^2)/(a^2+b^2) ...1 Given, 0 0 ⇒ a>b ...2 Now, as, C=90° ⇒ ∆ABC is a right-angled triangle whose hypotenuse is c. Hence, c>a>b (from 2)

Q67 / 160 · TS EAMCET 2021 (05 Aug Shift 2) · Mathematics

In a $∆ A B C,$ if $∠ C = 90 °, \frac{a^{2} + b^{2}}{a^{2} - b^{2}} \sin (A - B) = 1$ and $0 < B < 45 °,$ then

A
$a > b > c$
Correct answer
B
$c > a > b$
C
$c > b > a$
D
$a < b < c$

Solution

We have

$\frac{a^{2} + b^{2}}{a^{2} - b^{2}} \sin (A - B) = 1 \Rightarrow \sin (A - B) = \frac{a^{2} - b^{2}}{a^{2} + b^{2}} . . . (1)$

Given, $0 < B < 45 °$ and $∠ C = 90 °$

So, $0 < A - B < 90 °$ $\Rightarrow 0 < \sin (A - B) < 1 \Rightarrow 0 < \frac{a^{2} - b^{2}}{a^{2} + b^{2}} < 1$ (from $(1)$ )

$a^{2} - b^{2} > 0 \Rightarrow a > b . . . (2)$

Now, as, $C = 90 ° \Rightarrow ∆ A B C$ is a right-angled triangle whose hypotenuse is $c$ .

Hence, $c > a > b$ (from $(2)$ )

Hence, the correct answer is option: 'B'.

In a ∆ABC, if ∠C=90°, a2+b2a2-b2sin(A-B)=1 and 0<B<45°, then

Solution

Relevant Questions

In a $∆ A B C,$ if $∠ C = 90 °, \frac{a^{2} + b^{2}}{a^{2} - b^{2}} \sin (A - B) = 1$ and $0 < B < 45 °,$ then