Determining the Type of Polar Coordinate Triangle from Vertices

The problem asks us to identify the type of triangle given its vertices in polar coordinates. The vertices are provided as \((0, 0)\), \(\left( {3,\frac{\pi }{2}} \right)\), and \(\left( {3,\frac{\pi }{6}} \right)\). To determine the type of triangle (like equilateral, isosceles, or right-angled), we need to find the lengths of its sides. We can do this by converting the polar coordinates of the triangle vertices to Cartesian coordinates and then using the standard distance formula, or by using the polar distance formula directly.

Converting Polar Coordinates to Cartesian Coordinates

A point with polar coordinates \((r, \theta)\) can be converted to Cartesian coordinates \((x, y)\) using the formulas:

\(x = r \cos \theta\)

\(y = r \sin \theta\)

Let the vertices be A, B, and C:

• Vertex A: \((r_A, \theta_A) = (0, 0)\)
• Vertex B: \((r_B, \theta_B) = \left( {3,\frac{\pi }{2}} \right)\)
• Vertex C: \((r_C, \theta_C) = \left( {3,\frac{\pi }{6}} \right)\)

Converting these triangle vertices to Cartesian coordinates:

• A: \(x_A = 0 \cos(0) = 0\), \(y_A = 0 \sin(0) = 0\). So, A is at \((0, 0)\).
• B: \(x_B = 3 \cos\left(\frac{\pi}{2}\right) = 3 \times 0 = 0\), \(y_B = 3 \sin\left(\frac{\pi}{2}\right) = 3 \times 1 = 3\). So, B is at \((0, 3)\).
• C: \(x_C = 3 \cos\left(\frac{\pi}{6}\right) = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}\), \(y_C = 3 \sin\left(\frac{\pi}{6}\right) = 3 \times \frac{1}{2} = \frac{3}{2}\). So, C is at \(\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)\).

The triangle vertices in Cartesian coordinates are A\((0, 0)\), B\((0, 3)\), and C\(\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)\). Now we use the Cartesian distance formula to find the lengths of the sides of the Polar Coordinate Triangle.

Calculating Side Lengths Using the Distance Formula

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\). We apply this distance formula to each pair of triangle vertices.

Length of AB:

\(AB = \sqrt{(0-0)^2 + (3-0)^2} = \sqrt{0^2 + 3^2} = \sqrt{0 + 9} = \sqrt{9} = 3\)

Length of AC:

\(AC = \sqrt{\left(\frac{3\sqrt{3}}{2}-0\right)^2 + \left(\frac{3}{2}-0\right)^2} = \sqrt{\left(\frac{3\sqrt{3}}{2}\right)^2 + \left(\frac{3}{2}\right)^2}\)

\(AC = \sqrt{\frac{(3\sqrt{3})^2}{2^2} + \frac{3^2}{2^2}} = \sqrt{\frac{9 \times 3}{4} + \frac{9}{4}} = \sqrt{\frac{27}{4} + \frac{9}{4}}\)

\(AC = \sqrt{\frac{27+9}{4}} = \sqrt{\frac{36}{4}} = \sqrt{9} = 3\)

Length of BC:

\(BC = \sqrt{\left(\frac{3\sqrt{3}}{2}-0\right)^2 + \left(\frac{3}{2}-3\right)^2} = \sqrt{\left(\frac{3\sqrt{3}}{2}\right)^2 + \left(-\frac{3}{2}\right)^2}\)

\(BC = \sqrt{\frac{27}{4} + \frac{9}{4}} = \sqrt{\frac{36}{4}} = \sqrt{9} = 3\)

Analysing Side Lengths to Identify the Triangle Type

We have calculated the lengths of all three sides of the Polar Coordinate Triangle:

• AB = 3
• AC = 3
• BC = 3

Since all three sides of the triangle have equal length (3 units), the triangle is an equilateral triangle.

This analysis confirms that the triangle formed by the given polar coordinates is an equilateral triangle. The calculation involved converting the polar coordinates to Cartesian coordinates and then applying the distance formula.