## you have an isosceles triangle with two sides of length 2, and the included angle is π/6

so the area is 1/2 * 2*2 * sin π/6 = 1

Two points on the polar plane are given as p1 = (−2,3π/2 ) and p2 = (2,2π/3). Please find the triangular area formed by P1, P2, and the pole.

oobleck
answered

2 years ago

2 years ago

so the area is 1/2 * 2*2 * sin π/6 = 1

help
answered

2 years ago

2 years ago

help please
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2 years ago

2 years ago

Explain Bot
answered

11 months ago

11 months ago

To find the triangular area formed by the given points, we can use the formula for the area of a triangle in the polar coordinate system. The formula states that the area of a triangle with vertices (r1, θ1), (r2, θ2), and (r3, θ3) is given by:

Area = (1/2) * |(r1 * r2 * sin(θ2 - θ1)) + (r2 * r3 * sin(θ3 - θ2)) + (r3 * r1 * sin(θ1 - θ3))|

In this case, we need to find the area of the triangle formed by p1 = (-2, 3π/2), p2 = (2, 2π/3), and the origin (pole).

Let's calculate the area step by step:

1. Calculate the differences in angles between the points:

θ2 - θ1 = (2π/3) - (3π/2)

= (4π/6) - (9π/6)

= -5π/6

θ3 - θ2 = 0 - (2π/3)

= (-2π/3)

θ1 - θ3 = (3π/2) - 0

= (9π/6)

Note: The differences should always be between -π and π.

2. Substitute the values into the formula and calculate:

Area = (1/2) * |(-2 * 2 * sin(-5π/6)) + (2 * 2 * sin(-2π/3)) + (2 * -2 * sin(9π/6))|

Using the trigonometric identities sin(-x) = -sin(x) and sin(x + π) = -sin(x), we can simplify:

Area = (1/2) * |-2 * 2 * sin(5π/6) + 2 * 2 * sin(2π/3) - 2 * -2 * sin(π/6)|

Area = (1/2) * |4 * (sqrt(3)/2) + 4 * sqrt(3)/2 + 2 * (1/2)|

Area = (1/2) * |4sqrt(3) + 6|

Area = (1/2) * (4sqrt(3) + 6)

Area = 2sqrt(3) + 3

So, the triangular area formed by p1, p2, and the pole is 2sqrt(3) + 3 square units.