Solveeit Logo
Login

Question

Question: In ambiguous case if a, b and A are given and if there are two possible values of third side, are \(...

In ambiguous case if a, b and A are given and if there are two possible values of third side, are c1c_{1} and c2c_{2}, then

A

c1−c2=2(a2+b2sin⁡2A)c_{1} - c_{2} = 2\sqrt{(a^{2} + b^{2}\sin^{2}A)}

B

c1−c2=2(a2−b2sin⁡2A)c_{1} - c_{2} = 2\sqrt{(a^{2} - b^{2}\sin^{2}A)}

C

c1−c2=4(a2+b2sin⁡2A)c_{1} - c_{2} = 4\sqrt{(a^{2} + b^{2}\sin^{2}A)}

D

c1−c2=3(a2−b2sin⁡2A)c_{1} - c_{2} = 3\sqrt{(a^{2} - b^{2}\sin^{2}A)}

Answer

c1−c2=2(a2−b2sin⁡2A)c_{1} - c_{2} = 2\sqrt{(a^{2} - b^{2}\sin^{2}A)}

Explanation

Solution

cos⁡A=b2+c2−a22bc\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc} or c2−(2bcos⁡A)c+(b2−a2)=0c^{2} - (2b\cos A)c + (b^{2} - a^{2}) = 0

Which is quadratic equation in c. Let there be two roots, c1c_{1} and c2c_{2} of above quadratic equation then c1+c2=2bcos⁡Ac_{1} + c_{2} = 2b\cos A and c1c2=b2−a2c_{1}c_{2} = b^{2} - a^{2}

∴\therefore c1−c2=[(c1+c2)2−4c1c2]c_{1} - c_{2} = \sqrt{\lbrack(c_{1} + c_{2})^{2} - 4c_{1}c_{2}\rbrack} = [(2bcos⁡A)2−4(b2−a2)\sqrt{\lbrack(2b\cos A)^{2} - 4(b^{2} - a^{2})}

= [4a2−4b2(1−cos⁡2A)]\sqrt{\lbrack 4a^{2} - 4b^{2}(1 - \cos^{2}A)\rbrack}= 2(a2−b2sin⁡2A)2\sqrt{(a^{2} - b^{2}\sin^{2}A)}.