Question

Find the area of the triangle formed by the coordinated of whose angular points are $ \left( { - 3, - {{30}^ \circ }} \right),\left( {5,{{150}^ \circ }} \right) $ and $ \left( {7,{{210}^ \circ }} \right) $

Answer 1

Hint: This problem deals with the area of the triangle. Here, given the angular points, we have to convert these coordinates into polar coordinates and then find the area of the triangle from the obtained polar coordinates.
The area of the triangle which is obtained from the polar coordinates formula is used here:
 \[ \Rightarrow \dfrac{1}{2}\left| {{x_1}{y_2} - {x_2}{y_1} + {x_2}{y_3} - {x_3}{y_2} + {x_3}{y_1} - {x_1}{y_3}} \right|\]
Where the points $ \left( {{x_1},{y_1}} \right) $ , $ \left( {{x_2},{y_2}} \right) $ , $ \left( {{x_3},{y_3}} \right) $ are the vertices of the triangle.

Complete step-by-step answer:
Given three angular points which are also called as polar coordinates which are: $ \left( { - 3, - {{30}^ \circ }} \right),\left( {5,{{150}^ \circ }} \right) $ and $ \left( {7,{{210}^ \circ }} \right) $ .
Consider the first angular point $ \left( { - 3, - {{30}^ \circ }} \right) $ ,
Here $ r = - 3 $ and $ \theta = - {30^ \circ } $ .
Let the x and y coordinate of this angular point be $ A = \left( {{x_1},{y_1}} \right) $
Then the x-coordinate of the polar coordinate is given by:
 $ \Rightarrow x = r\cos \theta $
 $ \Rightarrow x = \left( { - 3} \right)\cos \left( { - {{30}^ \circ }} \right) $
As we know that $ \Rightarrow \cos \left( { - {{30}^ \circ }} \right) = \cos \left( {{{30}^ \circ }} \right) $
 $ \Rightarrow x = \left( { - 3} \right)\cos \left( {{{30}^ \circ }} \right) $
As the value of $ \cos \left( {{{30}^ \circ }} \right) = \dfrac{{\sqrt 3 }}{2} $ ;
 $ \Rightarrow x = \left( { - 3} \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right) $
 $ \Rightarrow x = \dfrac{{ - 3\sqrt 3 }}{2} $
The y-coordinate of the polar coordinate is given by:
 $ \Rightarrow y = r\sin \theta $
 $ \Rightarrow y = \left( { - 3} \right)\sin \left( { - {{30}^ \circ }} \right) $
As the value of $ \sin \left( { - {{30}^ \circ }} \right) = \dfrac{{ - 1}}{2} $ ;
 \[ \Rightarrow y = \left( { - 3} \right)\left( {\dfrac{{ - 1}}{2}} \right)\]
 $ \Rightarrow y = \dfrac{3}{2} $
 $ \therefore A = \left( {\dfrac{{ - 3\sqrt 3 }}{2},\dfrac{3}{2}} \right) $
Now consider the second angular point $ \left( {5,{{150}^ \circ }} \right) $ ,
Here $ r = 5 $ and $ \theta = {150^ \circ } $ .
Let the x and y coordinate of this angular point be $ B = \left( {{x_2},{y_2}} \right) $
Then the x-coordinate of the polar coordinate is given by:
 $ \Rightarrow x = r\cos \theta $
 $ \Rightarrow x = \left( 5 \right)\cos \left( {{{150}^ \circ }} \right) $
As the value of $ \cos \left( {{{150}^ \circ }} \right) = \dfrac{{ - \sqrt 3 }}{2} $ ;
 $ \Rightarrow x = \left( 5 \right)\left( {\dfrac{{ - \sqrt 3 }}{2}} \right) $
 $ \Rightarrow x = \dfrac{{ - 5\sqrt 3 }}{2} $
The y-coordinate of the polar coordinate is given by:
 $ \Rightarrow y = r\sin \theta $
 $ \Rightarrow y = \left( 5 \right)\sin \left( {{{150}^ \circ }} \right) $
As the value of $ \sin \left( {{{150}^ \circ }} \right) = \dfrac{1}{2} $ ;
 $ \Rightarrow y = \left( 5 \right)\left( {\dfrac{1}{2}} \right) $
 $ \Rightarrow y = \dfrac{5}{2} $
 $ \therefore B = \left( {\dfrac{{ - 5\sqrt 3 }}{2},\dfrac{5}{2}} \right) $
Now consider the third angular point $ \left( {7,{{210}^ \circ }} \right) $ ,
Here $ r = 7 $ and $ \theta = {210^ \circ } $ .
Let the x and y coordinate of this angular point be $ C = \left( {{x_3},{y_3}} \right) $
Then the x-coordinate of the polar coordinate is given by:
 $ \Rightarrow x = r\cos \theta $
 $ \Rightarrow x = \left( 7 \right)\cos \left( {{{210}^ \circ }} \right) $
As the value of $ \cos \left( {{{210}^ \circ }} \right) = \dfrac{{ - \sqrt 3 }}{2} $ ;
 $ \Rightarrow x = \left( 7 \right)\left( {\dfrac{{ - \sqrt 3 }}{2}} \right) $
 $ \Rightarrow x = \dfrac{{ - 7\sqrt 3 }}{2} $
The y-coordinate of the polar coordinate is given by:
 $ \Rightarrow y = r\sin \theta $
 $ \Rightarrow y = \left( 7 \right)\sin \left( {{{210}^ \circ }} \right) $
 $ \Rightarrow y = \left( 7 \right)\left( {\dfrac{{ - 1}}{2}} \right) $
As the value of $ \sin \left( {{{210}^ \circ }} \right) = \dfrac{{ - 1}}{2} $ ;
 $ \Rightarrow y = \dfrac{{ - 7}}{2} $
 $ \therefore C = \left( {\dfrac{{ - 7\sqrt 3 }}{2},\dfrac{{ - 7}}{2}} \right) $
So the vertices of the triangle are : $ A\left( {\dfrac{{ - 3\sqrt 3 }}{2},\dfrac{3}{2}} \right) $ , $ B\left( {\dfrac{{ - 5\sqrt 3 }}{2},\dfrac{5}{2}} \right) $ and $ C\left( {\dfrac{{ - 7\sqrt 3 }}{2},\dfrac{{ - 7}}{2}} \right) $ .
The area of the triangle is given by:
 \[ \Rightarrow \Delta ABC = \dfrac{1}{2}\left| {{x_1}{y_2} - {x_2}{y_1} + {x_2}{y_3} - {x_3}{y_2} + {x_3}{y_1} - {x_1}{y_3}} \right|\]
 \[ \Rightarrow \Delta ABC = \dfrac{1}{2}\left| {\left( {\dfrac{{ - 3\sqrt 3 }}{2}} \right)\left( {\dfrac{5}{2}} \right) - \left( {\dfrac{{ - 5\sqrt 3 }}{2}} \right)\left( {\dfrac{3}{2}} \right) + \left( {\dfrac{{ - 5\sqrt 3 }}{2}} \right)\left( {\dfrac{{ - 7}}{2}} \right) - \left( {\dfrac{{ - 7\sqrt 3 }}{2}} \right)\left( {\dfrac{5}{2}} \right) + \left( {\dfrac{{ - 7\sqrt 3 }}{2}} \right)\left( {\dfrac{3}{2}} \right) - \left( {\dfrac{{ - 3\sqrt 3 }}{2}} \right)\left( {\dfrac{{ - 7}}{2}} \right)} \right|\]
Multiplying the terms inside the modulus as given by:
 \[ \Rightarrow \Delta ABC = \dfrac{1}{2}\left| {\left( {\dfrac{{ - 15\sqrt 3 }}{4}} \right) - \left( {\dfrac{{ - 15\sqrt 3 }}{4}} \right) + \left( {\dfrac{{35\sqrt 3 }}{4}} \right) - \left( {\dfrac{{ - 35\sqrt 3 }}{4}} \right) + \left( {\dfrac{{ - 21\sqrt 3 }}{4}} \right) - \left( {\dfrac{{21\sqrt 3 }}{4}} \right)} \right|\]
 \[ \Rightarrow \Delta ABC = \dfrac{1}{2}\left| {\dfrac{{ - 15\sqrt 3 }}{4} + \dfrac{{15\sqrt 3 }}{4} + \dfrac{{35\sqrt 3 }}{4} + \dfrac{{35\sqrt 3 }}{4} - \dfrac{{21\sqrt 3 }}{4} - \dfrac{{21\sqrt 3 }}{4}} \right|\]
Here the terms \[\dfrac{{15\sqrt 3 }}{4}\] and \[\dfrac{{ - 15\sqrt 3 }}{4}\] gets cancelled, as simplified below:
 \[ \Rightarrow \Delta ABC = \dfrac{1}{2}\left| {\dfrac{{35\sqrt 3 }}{2} - \dfrac{{21\sqrt 3 }}{2}} \right|\]
 \[ \Rightarrow \Delta ABC = \dfrac{1}{2}\left| {\dfrac{{14\sqrt 3 }}{2}} \right|\]
 $ \therefore \Delta ABC = \dfrac{{7\sqrt 3 }}{2} $
Hence the area of the triangle is $ \dfrac{{7\sqrt 3 }}{2} $ sq. units.

Final answer: The area of the triangle formed by the coordinated of whose angular points are $ \left( { - 3, - {{30}^ \circ }} \right),\left( {5,{{150}^ \circ }} \right) $ and $ \left( {7,{{210}^ \circ }} \right) $ is $ \dfrac{{7\sqrt 3 }}{2} $ sq. units.

Note:
Please note that while solving the problem, we should understand that only the cosine and secant trigonometric ratios of negative angles is positive, all the other trigonometric ratios of negative angles are negative. Which is given by:
 $ \Rightarrow \sin \left( { - \theta } \right) = - \sin \theta $
 $ \Rightarrow \cos \left( { - \theta } \right) = \cos \theta $
 $ \Rightarrow \tan \left( { - \theta } \right) = - \tan \theta $
 $ \Rightarrow \cot \left( { - \theta } \right) = - \cot \theta $
 $ \Rightarrow \sec \left( { - \theta } \right) = \sec \theta $
 $ \Rightarrow \cos ec\left( { - \theta } \right) = - \cos ec\theta $