সৃজনশীলঃ ‘…’ আদমজী ক্যান্টনমেন্ট কলেজ ২০২৪ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $\theta =C-{40}^{\circ }$ এবং $C={60}^{\circ }$

$\therefore \theta ={60}^{\circ }-{40}^{\circ }={20}^{\circ }$

এখন, $\tan {20}^{\circ }\tan {40}^{\circ }\tan {80}^{\circ }=\tan {20}^{\circ }\tan ({60}^{\circ }-{20}^{\circ })\tan ({60}^{\circ }+{20}^{\circ })$

$=\tan {20}^{\circ }\times \frac{\tan {60}^{\circ }-\tan {20}^{\circ }}{1+\tan {60}^{\circ }\cdot \tan {20}^{\circ }}\times \frac{\tan {60}^{\circ }+\tan {20}^{\circ }}{1-\tan {60}^{\circ }\cdot \tan {20}^{\circ }}$

$=\tan {20}^{\circ }\times \frac{\sqrt{3}-\tan {20}^{\circ }}{1+\sqrt{3}\tan {20}^{\circ }}\times \frac{\sqrt{3}+\tan {20}^{\circ }}{1-\sqrt{3}\tan {20}^{\circ }}$

$=\tan {20}^{\circ }\times \frac{(\sqrt{3}{)}^2-{\tan }^2{20}^{\circ }}{1-{(\sqrt{3}\tan {20}^{\circ })}^2}=\tan {20}^{\circ }\times \frac{3-{\tan }^2{20}^{\circ }}{1-3{\tan }^2{20}^{\circ }}$

$=\frac{3\tan {20}^{\circ }-{\tan }^3{20}^{\circ }}{1-3{\tan }^2{20}^{\circ }}=\tan (3\times {20}^{\circ })=\tan {60}^{\circ }=\sqrt{3}$

বামপক্ষ $=\tan \theta \cdot \tan 2\theta \cdot \tan 3\theta \cdot \tan 4\theta$

$=\tan {20}^{\circ }\cdot \tan {40}^{\circ }\cdot \tan {60}^{\circ }\cdot \tan {80}^{\circ }$

$=\tan {20}^{\circ }\cdot \tan {40}^{\circ }\cdot \tan {80}^{\circ }\cdot \tan {60}^{\circ }$

$=\sqrt{3}\times \sqrt{3}=3=$ ডানপক্ষ

$\therefore \tan \theta \tan 2\theta \tan 3\theta \tan 4\theta =3$ (প্রমাণিত)

$a=\sqrt{3}+1,b=\sqrt{3}-1$ এবং $\angle C={60}^{\circ }$

আমরা জানি, $c^2=a^2+b^2-2\mathrm{ab}\cos C$

$=(\sqrt{3}+1{)}^2+(\sqrt{3}-1{)}^2-2(\sqrt{3}+1)(\sqrt{3}-1)\cos {60}^{\circ }$

$=2(3+1)-2(3-1)\frac{1}{2}=8-2=6$

$\therefore c=\sqrt{6}\text{ (Ans.)}$

ত্রিভুজের সাইন সূত্র হতে আমরা পাই,

$\sin {15}^{\circ }=\sin ({45}^{\circ }-{30}^{\circ })$

$=\sin {45}^{\circ }\cos {30}^{\circ }-\cos {45}^{\circ }\sin {30}^{\circ }$

$=\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\cdot \frac{1}{2}$

$=\frac{\sqrt{3}-1}{2\sqrt{2}}$

$\frac{b}{\sin B}=\frac{c}{\sin C}$

বা, $\sin B=\frac{b}{c}\sin C$

$=\frac{\sqrt{3}-1}{\sqrt{6}}\sin {60}^{\circ }$

$=\frac{\sqrt{3}-1}{\sqrt{3}\cdot \sqrt{2}}\cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}}$

$\therefore B={\sin }^{-1}\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)={15}^{\circ }\text{ (Ans.) }$

$\therefore A={180}^{\circ }-(B+C)={180}^{\circ }-{15}^{\circ }-{60}^{\circ }={105}^{\circ }\text{ (Ans.)}$