Question

In triangle $ABC,\angle A={30}^0,b=6$. Let $C{B}_1$ and $C{B}_2$ are least and greatest integer value of side $a$ for which two triangles can be formed. It is also given angle $B_1$ is obtuse and angle $B_2$ is acute angle. (All symbols used have usual meaning in a triangle)

• A. $\left|C{B}_1-C{B}_2\right|=1$
• B. $C{B}_1+C{B}_2=9$
• C. area of $\Delta B_{1}C{B}_2=6+\frac{3}{2}\sqrt{7}$
• D. area of $\Delta A{B}_{2}C=6+\frac{9}{2}\sqrt{3}$

Answer 1

The correct option is D area of $\Delta A{B}_{2}C=6+\frac{9}{2}\sqrt{3}$

$\begin{array}{c}\cos {30}^0=\frac{c^2+b^2-a^2}{2bc}\\ \frac{\sqrt{3}}{2}=\frac{c^2+36-a^2}{12c}\\ 6\sqrt{3}c=c^2+36-a^2\\ differentiating,\\ 6\sqrt{3}dc=2cdc+2ada\\ 3\sqrt{3}dc=cdc+ada\\ ada=\left(c-3\sqrt{3}\right)dc\\ \frac{da}{dc}=\frac{\left(c-3\sqrt{3}\right)}{a}=0\\ c=3\sqrt{3}\\ 6\sqrt{3}c=c^2+36-a^2\\ a^2=27+36-6\sqrt{3}\times 3\times \sqrt{3}\\ a^2=63-54\\ a=3\\ fromthis\\ \left|C{B}_1-C{B}_2\right|=|5-4|=1\\ C{B}_1+C{B}_2=9\\ areaof\Delta B_{1}C{B}_2=6+\frac{3}{2}\sqrt{7}\\ areaof\Delta A{B}_{2}C=6+\frac{9}{2}\sqrt{3}\end{array}$