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In a polar coordinate system ,O is the pole.The polar coordinates of A and B are(6,15) and (6,255) respectively.P is a moving point in the system such that PA=PB.

If Q is a point lying on P such that OABQ is a rhombus , find polar coordinates of Q

Guest Mar 5, 2015

Best Answer 

 #4
avatar+27057 
+5

Nice, elegant reasoning Bertie!

.

Alan  Mar 5, 2015
 #1
avatar+27057 
+5

 Rhombus

 Image

.

Alan  Mar 5, 2015
 #2
avatar+20025 
+5

In a polar coordinate system ,O is the pole.The polar coordinates of A and B are(6,15) and (6,255) respectively.P is a moving point in the system such that PA=PB.If Q is a point lying on P such that OABQ is a rhombus , find polar coordinates of Q

$$\small{\text{$
\vec{Q}=\vec{A}+\vec{B}
$, because $|\vec{A}| = |\vec{B}| = 6 = r \quad$
Q lies on the bisectors of an angle by O
}}\\\\
\small{\text{
$\vec{A}=
\left(
\begin{array}{c}r\cdot \cos{(15)}\\r\cdot \sin{(15)}\end{array}
\right)
=
\left(
\begin{array}{c}6\cdot \cos{(15)}\\6\cdot \sin{(15)}\end{array}
\right)$
}}\\
\small{\text{
$\vec{B}=
\left(
\begin{array}{c}r\cdot \cos{(255)}\\r\cdot \sin{(255)}\end{array}
\right)
=
\left(
\begin{array}{c}6\cdot \cos{(255)}\\6\cdot \sin{(255)}\end{array}
\right)
$
}}\\\\
\small{\text{$
\vec{Q}=\vec{A}+\vec{B}
=
\left(
\begin{array}{c}6\cdot \cos{(15)}\\6\cdot \sin{(15)}\end{array}
\right)
+
\left(
\begin{array}{c}6\cdot \cos{(255)}\\6\cdot \sin{(255)}\end{array}
\right)
=
\left(
\begin{array}{c}
6\cdot \cos{(15)} + 6\cdot \cos{(255)}
\\ 6\cdot \sin{(15)} + 6\cdot \sin{(255)}
\end{array}
\right)
$}}\\\\
\small{\text{$
=
\left(
\begin{array}{c} 4.24264068712 \\ -4.24264068712 \end{array}
\right)
=
\left(
\begin{array}{c} \sqrt{18} \\ - \sqrt{18} \end{array}
\right)
$}}$$

polar coordinates of Q

$$\small{\text{
$
\vec{Q}= \left(\begin{array}{c} Q_x =\sqrt{18}\\ Q_y=-\sqrt{18} \end{arry} \right)
\qquad r = \sqrt{ Q_x^2 + Q_y^2 } = \sqrt{18+18} = 6
$
}}\\\\
\small{\text{
$
\tan{(\varphi)} = \dfrac{Q_y}{ Q_x} = \dfrac{-\sqrt{18}}{\sqrt{18}} = -1 \qquad\varphi = -45\ensurement{^{\circ}} $ or $ \varphi = 315\ensurement{^{\circ}} $
}}\\\\
\vec{Q}=(6, 315\ensurement{^{\circ}} )$$

heureka  Mar 5, 2015
 #3
avatar+890 
+5

I'm going to cheat slightly and refer to Alan's rather nice diagram. (I drew it on paper).

The angle that OB makes with the negative y-axis is 15 deg which means that the angle AOB is 120 deg, which in turn means that the angle OBQ is 60 deg.

OB and BQ being the same length implies that the triangle OBQ is equilateral in which case the length of OQ is 6 and it's at an angle of 45 deg to the negative y-axis.

That means that Q has polar co-ordinates (6, -45deg).

Bertie  Mar 5, 2015
 #4
avatar+27057 
+5
Best Answer

Nice, elegant reasoning Bertie!

.

Alan  Mar 5, 2015

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