Let r1 and r2 be positive real numbers and 0<θ2 − θ1<π. Prove that the area of the triangle with verticesrole="math" localid="1663155755932" (0, 0), (r1, θ1), and (r2, θ2) in the polar plane is r1r22sin(θ2−θ1)

Q. D - Calculus | StudyQuestionHub

Q. D

Question

Let $r_{1}$ and $r_{2}$ be positive real numbers and $0 < θ_{2} - θ_{1} < π$ . Prove that the area of the triangle with vertices $(0, 0), (r_{1}, θ_{1}), and (r_{2}, θ_{2})$ in the polar plane is $\frac{r_{1} r_{2}}{2} \sin (θ_{2} - θ_{1})$

Step-by-Step Solution

Verified

Answer

Hence, proved

1Step 1: Given information

The vertices of the triangle are:

$(0, 0), (r_{1}, θ_{1}), and (r_{2}, θ_{2})$

2Step 2: Area of the triangle

The area of the triangle is given by,

$A = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] sq . units$

Here,

$x_{1} = r_{1} \cos θ_{1}, y_{1} = r_{1} \sin θ_{1} x_{2} = r_{2} \cos θ_{2}, y_{2} = r_{2} \sin θ_{2} x_{3} = 0, y_{3} = 0$

Now,

$A = \frac{1}{2} [r_{1} \cos θ_{1} (r_{2} \sin θ_{2} - 0) + r_{2} \cos θ_{2} (0 - r_{1} \sin θ_{1}) + 0 (r_{1} \sin θ_{1} - r_{2} \sin θ_{2})] sq . units A = \frac{1}{2} [r_{1} r_{2} \cos θ_{1} \sin θ_{2} - r_{2} r_{1} \sin θ_{1} \cos θ_{2}] sq . units A = \frac{1}{2} [r_{1} r_{2} (\cos θ_{1} \sin θ_{2} - \sin θ_{1} \cos θ_{2})] sq . units A = \frac{1}{2} [r_{1} r_{2} (\sin (θ_{2} - θ_{1}))] sq . units A = \frac{r_{1} r_{2} \sin (θ_{2} - θ_{1})}{2} sq . units$

Hence, proved.

Q. C

Q. 1

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Q. C

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