from an observation point the angles of depression of two boats in line with this point are found to 30° and 45°. find the distance between the...

Sunday, November 23, 2014

Form a right triangle ABC with the given information, such that

AB = height of the point of observation = $\displaystyle{4000}{f{}}$

BC=distance from the foot of the point to the first boat= $\displaystyle{x}$ , $\displaystyle\angle$ ACB = $\displaystyle{{45}}^{{0}}$

Produce BC to D, such that $\displaystyle\angle$ ADB = $\displaystyle{{30}}^{{0}}$ , and DC =distance between two boats= $\displaystyle{d}{m}$

Now we evaluate these values by using trigonometric functions.

In $\displaystyle\delta$ ABC

$\displaystyle{{\tan{{45}}}}^{{0}}=\frac{{{A}{B}}}{{{B}{C}}}$

$\displaystyle{1}=\frac{{4000}}{{x}}$

$\displaystyle\therefore{x}={4000}$

In $\displaystyle\delta$ ADB

$\displaystyle{{\tan{{30}}}}^{{0}}=\frac{{4000}}{{{x}+{d}}}$

$\displaystyle\frac{{1}}{\sqrt{{3}}}=\frac{{4000}}{{{x}+{d}}}$

$\displaystyle{x}+{d}={4000}\sqrt{{3}}$

by substituting $\displaystyle{x}$ value, we get

$\displaystyle{4000}+{d}={4000}\sqrt{{3}}$

$\displaystyle{d}={4000}{\left(\sqrt{{3}}-{1}\right)}={2928.4}{f{}}$

$\displaystyle\therefore$ the boats are $\displaystyle{2928.4}{f{}}$ apart.

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