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A is a point on the $x$ axis and B is a point on the $y$ axis such that $(5,3)$ lies on the straight line passing through A and B . Given that OP is perpendicular to AB and $\angle BAO = \theta $ , Show that OP = $3 \ cos \theta + 5 \sin \theta $

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My attempt ,

$\sin \theta = \frac{OP}{OA} $

$OP = OA \sin \theta$

OA = $ 5 + ? $

The '?' is what I marked on the diagram as well. I'm not sure how do I get that .

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  • $\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$ – user Jan 26 '18 at 21:10
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$$\tan \theta = \frac{3}{OA-5} $$

$$OA-5=3\cot \theta$$

$$OA=3\cot \theta+5$$

Thus

$$\sin \theta= \frac{OP}{OA} $$

$$OP=OA\sin \theta =3\cos \theta+5\sin \theta$$

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  • $\begingroup$ Isn't $\tan \theta = \frac{OP}{OA} $ ? How did you get there $\endgroup$ – user175089 Dec 11 '17 at 7:18
  • $\begingroup$ $$\tan \theta= \frac{OP}{AP} $$ I’ve applied this at the little right triangle in A $\endgroup$ – user Dec 11 '17 at 7:23
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\begin{align} |OA|&=5+3\cot\theta ,\\ |OB|&=|OA|\tan\theta ,\\ |OP|&=|OB|\cos\theta =(5+3\cot\theta)\tan\theta \cos\theta =(5+3\cot\theta)\sin\theta =5\sin\theta+3\cos\theta . \end{align}

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The equation of $AB$

$$\tan(180^\circ-\theta)=\dfrac{y-3}{x-5}$$

$$\iff\dfrac x{\dfrac{3\cos\theta+4\sin\theta}{\sin\theta}}+\dfrac y{\dfrac{3\cos\theta+4\sin\theta}{\cos\theta}}=1$$

$OA=\dfrac{3\cos\theta+4\sin\theta}{\sin\theta}$

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The normal form of the equation of a straight line is $$ x cos α + y sin α - p = 0 $$
where $\alpha$ is the angle from the x-axis to the perpendicular from the origin to the line and p is the length of the perpendicular from origin to the line.

So, here, $\alpha=90-\theta$

So, put in the equation and also the length of the perpendicular to the line from the origin=$OP$ and also $(5,3)$ lies on the line. So, $$ 5 \cos (90-\theta) + 3 \sin (90-\theta) - p = 0 $$ $$\implies 5\sin \theta +3 \cos \theta=OP$$

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