Chapter 10: Q. D (page 777)

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 verticesrole="math" localid="1663155755932" $(0, 0), (r_{1}, θ_{1}), and (r_{2}, θ_{2})$ in the polar plane is $\frac{r_{1} r_{2}}{2} \sin (θ_{2} - θ_{1})$

Short Answer

Expert verified

Hence, proved

Step by step solution

Given information

The vertices of the triangle are:

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

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.

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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 verticesrole="math" localid="1663155755932" $(0, 0), (r_{1}, θ_{1}), and (r_{2}, θ_{2})$ in the polar plane is $\frac{r_{1} r_{2}}{2} \sin (θ_{2} - θ_{1})$

Short Answer

Step by step solution

Given information

Area of the triangle

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Let r1and r2be positive real numbers and0<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 isr1r22sin(θ2−θ1)

Short Answer

Step by step solution

Given information

Area of the triangle

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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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 verticesrole="math" localid="1663155755932" $(0, 0), (r_{1}, θ_{1}), and (r_{2}, θ_{2})$ in the polar plane is $\frac{r_{1} r_{2}}{2} \sin (θ_{2} - θ_{1})$