Trigonometric Equations Question 300
Question: If in triangle $ ABC,\frac{a^{2}-b^{2}}{a^{2}+b^{2}}=\frac{\sin (A-B)}{\sin (A+B)} $ , then the triangle is
[Roorkee 1987]
Options:
A) Right angled
B) Isosceles
C) Right angled or isosecles
D) Right angled isosecles
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \frac{\sin (A-B)}{\sin (A+B)}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}} $
Þ $ \frac{\sin (A-B)}{\sin (A+B)}=\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{{{\sin }^{2}}A+{{\sin }^{2}}B} $
Þ $ \frac{\sin (A-B)}{\sin C}=\frac{\sin (A-B)\sin (A+B)}{{{\sin }^{2}}A+{{\sin }^{2}}B} $
Þ $ \sin (A-B)[ \frac{1}{\sin C}-\frac{\sin C}{{{\sin }^{2}}A+{{\sin }^{2}}B} ]=0 $ Either $ \sin (A-B)=0\Rightarrow A=B $ i.e. isosceles or $ {{\sin }^{2}}A+{{\sin }^{2}}B={{\sin }^{2}}C $ or $ a^{2}+b^{2}=c^{2} $ i.e., right angled triangle.
BETA
