Question

find the centroid of the triangle whose vertices are ( cos0°, sin90°), (tan45°, cot45°) and (sec45°, cot6o°)​

Answer 1

Given: Vertices are ( cos0°, sin90°), (tan45°, cot45°) and (sec45°, cot6o°)​.

To find: The centroid of the triangle.

Solution:

• Let vertices are A( cos0°, sin90°), B(tan45°, cot45°) and C(sec45°, cot6o°)​.

• So we have given the vertices of the triangle in trigonometric form, so lets convert them to the numeric terms.

                  A = (1, 1), B = (1, 1), C = (√2, 1/√3)

• We know the formula of centroid, that is:

                  x1 + x2 + x3/3, y1 + y2 + y3/3

• Putting values in the formula we get:

                 ( 1+1+√2 ) /3 ,  ( 1+1+1/√3 ) / 3

                 ( 2+√2 ) / 3, (1 + 2√3) / 3√3

Answer:

                So, the centroid of the triangle whose vertices are ( cos0°, sin90°), (tan45°, cot45°) and (sec45°, cot6o°)​  is { ( 2+√2 ) / 3, (1 + 2√3) / 3√3 }