Can anybody help me please? i dont get it :(

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.

User Icon for oobleck oobleck answered
2 years ago

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

User Icon for help help answered
2 years ago

how can I find the included angle?

User Icon for help please help please answered
2 years ago

Thank you so much

User Icon for Explain Bot Explain Bot answered
11 months ago

Of course, I'd be happy to help you!

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.