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3 years ago

I need help on this on my bros

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

6 0

Step-by-step explanation:

By 30°-60°-90° triangle theorem:

$15 = \frac{1}{2} \times c.. (side\:opposite \: to \:30°)\\ c = 15 \times 2 \\ \\ \huge \red{ \boxed{c = 30 \: m}} \\ \\ b = \frac{ \sqrt{3} }{2} \times c.. (side\:opposite \: to \:60°) \\ \\ b = \frac{ \sqrt{3} }{2} \times 30 \\ \\ \huge \purple{ \boxed{b = 15 \sqrt{3} \: m}}$

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$\large\begin{array}{l}\\\\ \textsf{This is a complex number in }\mathsf{a+bi}\textsf{ form:}\\\\ \mathsf{z=-6+6\sqrt{3}\,i}\\\\ \textsf{where }\mathsf{a=-6~~and~~b=6\sqrt{3}.} \end{array}$

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$\large\begin{array}{l}\bullet~~\textsf{2}\mathsf{^{nd}}\textsf{ step: Find the argument }\mathsf{\theta}\textsf{ of z:}\\\\ \mathsf{arg(z)=\theta}\\\\ \textsf{is an angle that satisfies the following conditions:}\\\\ \begin{cases} \mathsf{cos\,\theta=\dfrac{a}{|z|}=\dfrac{a}{\sqrt{a^2+b^2}}}\\\\ \mathsf{sin\,\theta=\dfrac{b}{|z|}=\dfrac{b}{\sqrt{a^2+b^2}}} \end{cases}\qquad\quad-\pi$

$\large\begin{array}{l}\textsf{For }\mathsf{z=-6+6\sqrt{3}\,i,}\\\\ \begin{cases} \mathsf{cos\,\theta=\dfrac{-6}{12}=-\,\dfrac{1}{2}}\\\\ \mathsf{sin\,\theta=\dfrac{6\sqrt{3}}{12}=\dfrac{\sqrt{3}}{2}} \end{cases}\\\\\\ \textsf{Since}\\\\ \mathsf{cos\,\theta0,}\\\\ \theta\textsf{ lies in the 2}\mathsf{^{nd}}\textsf{ quadrant.}\\\\\\\textsf{Therefore,}\\\\ \mathsf{\theta=\dfrac{2\pi}{3}~rad~~(120^\circ)} \end{array}$

$\large\begin{array}{l}\bullet~~\textsf{3}\mathsf{^{rd}}\textsf{ step: Express z in the trigonometric form:}\\\\ \begin{array}{rcl} \mathsf{z}&\!\!=\!\!&\mathsf{|z|\cdot (cos\,\theta+i\,sin\,\theta)}\\\\ \mathsf{-6+6\sqrt{3}\,i}&\!\!=\!\!&\mathsf{12\cdot \left(cos\, \dfrac{2\pi}{3}+i\,sin\,\dfrac{2\pi}{3}\right)\qquad\quad\checkmark} \end{array} \end{array}$

If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2154167

$\large\textsf{I hope it helps.}$

Tags: complex number trigonometric form trig absolute value modulus argument angle

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I need help on this on my bros​

I need help on this on my bros