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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions4 Marks
Question
Solve the triangle in which $a=(\sqrt{3}+1), b=(\sqrt{3}-1)$ and $\angle C=60^{\circ}$.
✓

Answer

Given : $a=\sqrt{3}+1, b=\sqrt{3}-1$ and $\angle C=60^{\circ}$.
By cosine rule,$c^2=a^2+b^2-2 a b \cos C$
$\begin{aligned} & =(\sqrt{3}+1)^2+(\sqrt{3}-1)^2-2(\sqrt{3}+1)(\sqrt{3}-1) \cos 60^{\circ} \\ & =3+1+2 \sqrt{3}+3+1-2 \sqrt{3}-2(3-1)\left(\frac{1}{2}\right)\end{aligned}$
$\begin{aligned} & =8-2=6 \\ & \therefore c=\sqrt{6} \ldots[\because c>0)\end{aligned}$
By sine rule,
$\begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ \therefore & \frac{\sqrt{3}+1}{\sin A}=\frac{\sqrt{3}-1}{\sin B}=\frac{\sqrt{6}}{\sin 60^{\circ}}\end{aligned}$
$\begin{aligned} & \therefore \frac{\sqrt{3}+1}{\sin A}=\frac{\sqrt{3}-1}{\sin B}=\frac{\sqrt{6}}{(\sqrt{3} / 2)}=2 \sqrt{2} \\ & \therefore \sin A=\frac{\sqrt{3}+1}{2 \sqrt{2}} \text { and } \sin B=\frac{\sqrt{3}-1}{2 \sqrt{2}}\end{aligned}$
$\begin{aligned} & \therefore \sin A=\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}} \text { and } \sin B=\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}} \\ & \therefore \sin A=\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{1}{\sqrt{2}} \\ & \text { and } \sin B=\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}-\frac{1}{2} \times \frac{1}{\sqrt{2}}\end{aligned}$
∴ sin A = sin 60° cos 45° + cos 60° sin 45° and sin B = sin 60° cos 45° – cos 60° sin 45° ∴ sin A = sin (60° + 45°) – sin 105° and sin B = sin (60° – 45°) = sin 15° ∴ A = 105° and B = 15°
Hence, $\mathrm{A}=105^{\circ}, \mathrm{B} 15^{\circ}$ and $\mathrm{C}=\sqrt{6}$ units.

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