Question

If for a triangle ABC, a, b, A are given, then which of the following gives us two such triangles

• A. a < b sin A
• B. a $=$ b sin A
• C. a > b sin A and a < b
• D. a > b sin A and a > b

Answer 1

Answer: (C)

a > b sin A and a < b

Here $a,b,A$ are given. Thus using cosine rule,

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\Rightarrow c^2-\left(2b\cos A\right)c+(b^2-a^2)=0$

Now for two triangle to exist roots of above quadratic should be positive and real

$\Rightarrow b^2-a^2>0\Rightarrow a<b$

and $(2b\cos A{)}^2-4(b^2-a^2)>0$

$4b^2{\cos }^{2}A-4b^2+4a^2>0\Rightarrow 4a^2>4b^2(1-{\cos }^{2}A)\Rightarrow a^2>b^{2}si{n}^{2}A$

$\Rightarrow b\sin A<a$