JEE Main 2017 — Mathematics Trigonometry

$$\mathrm{cos}⁡ { 60}^{o} =\frac{4+25-{c}^{2}}{2.2.5}$$cos⁡60o=2.2.54+25−c2​

$$\Rightarrow 10=29-{c}^{2}$$⇒10=29−c2

$$\Rightarrow {c}^{2}=19$$⇒c2=19

$$\Rightarrow c=\sqrt{19}$$⇒c=19​

Now,

$$\mathrm{cos}120^{\circ}=\frac{ {a}^{2} +{b}^{2} -{c}^{2} }{ 2ab }$$cos120∘=2aba2+b2−c2​

$\Rightarrow$⇒ $-\frac{1}{2}=\frac{{a}^{2}+{b}^{2}-19}{2ab}$−21​=2aba2+b2−19​

$$\Rightarrow {a}^{2}+{b}^{2}-19=-ab$$⇒a2+b2−19=−ab

$$\Rightarrow {a}^{2}+{b}^{2}+ab=19$$⇒a2+b2+ab=19

Area of quadrilateral $=\frac{1}{2}\times 2\times 5\mathrm{sin}⁡60^{\circ}+\frac{1}{2}ab\mathrm{sin}⁡120^{\circ}=4\sqrt{3}$=21​×2×5sin⁡60∘+21​absin⁡120∘=43​

$$\Rightarrow \frac{5\sqrt{3}}{2}+\frac{ab\sqrt{3}}{4}=4\sqrt{3}$$⇒253​​+4ab3​​=43​

$$\Rightarrow \frac{ab}{4}=4-\frac{5}{2}=\frac{3}{2}$$⇒4ab​=4−25​=23​

$$\Rightarrow ab=6$$⇒ab=6

$$\Rightarrow {a}^{2}+{b}^{2}=13$$⇒a2+b2=13

$$\Rightarrow a=2, b=3$$⇒a=2,b=3

Perimeter

$$=2+5+2+3$$=2+5+2+3

$=12$=12 units