Question

In a triangle abc, a, b and a are given, b > a and c1, c2 are two possible values of the third side
c. If 1 and 2 are areas of two triangles with sides a, b, c1 and a, b, c2 then

Answer 1

If 1 and 2 are areas of two triangles with sides a, b, c1 and a, b, c2 then $\bold{|\Delta_1-\Delta_2|=b\sin A\sqrt{a^2-b^2\sin^2 A}}$

Solution:

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

$\bold{c(2b\cos A)=b^2+c^2-a^2}$

$\bold{b^2+c^2-a^2-c(2b\cos A)=0}$

The sum of the roots:

$\bold{c1+c2=2b\cos A}$

The product of the roots:

$\bold{c1.c2=b^2-a^2}$

The area of the triangle with side c1 is:

$\bold{\Delta_1=\frac{1}{2}bc1\sin A}$

The area of the triangle with side c2 is:

$\bold{\Delta_2=\frac{1}{2}bc2\sin A}$

Now, the arithmetic mean is:

$\bold{A.M=\frac{\Delta_1+\Delta_2}{2}}$

$\bold{A.M=\frac{1}{4}\times(c1+c2)b\sin A}$

On substituting the value of the sum of the roots, we get,

$\bold{A.M=\frac{1}{4}(2b\cos A)b\sin A}$

$\bold{[\because\sin A+\cos A=\sin 2A]}$

$\bold{\therefore A.M=\frac{1}{4}b^2\sin 2A}$

Now, the geometric mean is:

$\bold{G.M=\sqrt{\Delta_1\Delta_2}}$

$\bold{G.M=\sqrt{\frac{1}{4}\times c1c2\times b^2\sin^2A}}$

$\bold{G.M=\frac{1}{2}b\sin A\sqrt{c1.c2}}$

On substituting the value of product of the roots, we get,

$\bold{\therefore G.M=\frac{1}{2}b\sin A\sqrt{b^2-a^2}}$

Now, the harmonic mean is:

$\bold{H.M=\frac{2\Delta_1\Delta_2}{\Delta_1+\Delta_2}}$

$\bold{H.M=\frac{\frac{1}{4}\times c1c2\times b^2\sin^2 A}{\frac{1}{2}(c1+c2)b\sin A}}}$

on Substituting the values of sum and product of the roots, we get,

$\bold{H.M=\frac{b\sin A(b^2-a^2)}{2b\cos A}}$

$\bold{\therefore H.M=(\frac{b^2-a^2}{2})\tan A}$

Now,

$\bold{|\Delta_1-\Delta_2|^2=(\Delta_1+\Delta_2)^2-4\Delta_1\Delta_2}$

$\bold{|\Delta_1-\Delta_2|^2=\frac{1}{4}b^2\sin^2 2A-b^2\sin^2 A(b^2-A^2)}$

$\bold{|\Delta_1-\Delta_2|^2=b^2\sin^2A(a^2-b^2\sin^2 A)}$

$\bold{\therefore |\Delta_1-\Delta_2|=b\sin A\sqrt{a^2-b^2\sin^2 A}}$