TRIGONOMETRY,help!!! Draw a triangle ABC and label its sides. Let angle A=θ. Assume that a

Sunday, June 3, 2012

TRIGONOMETRY,help!!! Draw a triangle ABC and label its sides. Let angle A=θ. Assume that a

You need to use the law of cosines, such
that:

$\displaystyle{{\left({B}{C}\right)}}^{{2}}={{\left({A}{B}\right)}}^{{2}}+{{\left({A}{C}\right)}}^{{2}}-{2}{A}{B}\cdot{A}{C}\cdot{\cos{{\left({\hat{{A}}}\right)}}}$

$\displaystyle{{a}}^{{2}}={{x}}^{{2}}+{{b}}^{{2}}-{2}{x}\cdot{b}{\cos{\theta}}$

Moving the terms to one side
yields:

$\displaystyle{{x}}^{{2}}+{{b}}^{{2}}-{2}{x}\cdot{b}{\cos{\theta}}-{{a}}^{{2}}={0}$

You need to rearrange the terms, such
that:

$\displaystyle{{x}}^{{2}}-{2}{x}\cdot{b}{\cos{\theta}}+{{b}}^{{2}}-{{a}}^{{2}}={0}$

You need to use quadratic formula, such
that:

$\displaystyle{x}_{{{1},{2}}}={\left({2}{b}{\cos{\theta}}\pm\sqrt{{{4}{{b}}^{{2}}{{\cos}}^{{2}}\theta-{4}{{b}}^{{2}}+}}\right.}$
4a^2))/2

You need to notice that $\displaystyle{x}_{{{1},{2}}}$ exist if the radical is
valid, hence $\displaystyle{4}{{b}}^{{2}}{{\cos}}^{{2}}\theta-{4}{{b}}^{{2}}+{4}{{a}}^{{2}}\ge{0}$ such
that:

$\displaystyle{4}{{b}}^{{2}}{{\cos}}^{{2}}\theta-{4}{{b}}^{{2}}+{4}{{a}}^{{2}}\ge{0}$

$\displaystyle{4}{{b}}^{{2}}{{\cos}}^{{2}}\theta-{4}{\left({{b}}^{{2}}-{{a}}^{{2}}\right)}\ge{0}$

Since the problem provides that $\displaystyle{b}>{a}$ , hence $\displaystyle{{b}}^{{2}}>$
a^2 => b^2 - a^2 > 0

$\displaystyle{{\cos}}^{{2}}\theta\ge{\left({{b}}^{{2}}-\right.}$
a^2)/b^2 => cos theta >= +-(sqrt(b^2-a^2))/b

$\displaystyle{x}_{{{1},{2}}}={\left({2}{b}{\cos{\theta}}\pm{2}\sqrt{{{{b}}^{{2}}{{\cos}}^{{2}}\theta-{{b}}^{{2}}+}}\right.}$
a^2))/2

$\displaystyle{x}_{{{1},{2}}}={\left({b}{\cos{\theta}}\pm\sqrt{{{{b}}^{{2}}{\left({{\cos}}^{{2}}\theta-{1}\right)}+}}\right.}$
a^2))

$\displaystyle{x}_{{{1},{2}}}={\left({b}{\cos{\theta}}\pm\sqrt{{{{a}}^{{2}}-{{b}}^{{2}}{{\sin}}^{{2}}}}\right.}$
theta))

Hence, evaluating x, under the given
conditions, $\displaystyle{\cos{\theta}}\ge\pm\frac{{\sqrt{{{{b}}^{{2}}-{{a}}^{{2}}}}}}{{b}}$ , yields $\displaystyle{x}_{{{1},{2}}}={\left({b}{\cos{}}\right.}$
theta+- sqrt(a^2 - b^2 sin^2 theta)).

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Sunday, June 3, 2012