Question

$\theta =\arctan \left(\frac{30+18\sin \left(30^{\circ \:}\right)}{18\cos \left(30^{\circ \:}\right)-6}\right)$

solve

Answer 1

Explanation:

Explanation:

$\sf \to \: \theta =\arctan \left(\frac{30+18\sin \left(30^{\circ \:}\right)}{18\cos \left(30^{\circ \:}\right)-6}\right) \\ \\ \sf \to \: \theta =\arctan \left(\frac{30+18\sin \left(30^{\circ \:}\right)}{18\sin\left(60^{\circ \:}\right)-6}\right) \\ \\ \sf \to \red{ \theta =\arctan \left(\frac{5+ \left(30^{\circ \:}\right)}{\left(60^{\circ \:}\right)}\right)}$

Answer 2

$θ=\arctan \left(\frac{30+18\sin \left(30^{\circ \:}\right)}{18\cos \left(30^{\circ \:}\right)-6}\right)\quad :\quad θ=\arctan \left(\frac{13\left(3\sqrt{3}+2\right)}{23}\right)\quad \left(\mathrm{Decimal}:\quad θ=1.32971\dots \right)$

$\frac{30+18\sin \left(30^{\circ \:}\right)}{18\cos \left(30^{\circ \:}\right)-6}=\frac{13}{3\sqrt{3}-2}$

$θ=\arctan \left(\frac{13}{3\sqrt{3}-2}\right)$

$\mathrm{Simplify}$

$θ=\arctan \left(\frac{13\left(3\sqrt{3}+2\right)}{23}\right)$

Consider taking I as theta