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3 years ago

What is the perimeter of abc with vertices A(-2,9) B(7,-3) and C(-2,-3) in coordinate plane

2 answers:

6 0

For the answer to the question above,
A - C = ( -2, 9) - (- 2 -3)

= ( 0, 12) 

AC = 12 

B - C = (7 - 3) - ( - 2 -3) 

= (9, 0) 

BC = 9 

ABC is a right -angled triangle with AC as the hypotenuse 

AC = sqrt( 9^2 + 12^2) 

= sqrt(81 + 144) 

= srqt(225) 

= 15 

Perimeter = 12 + 9 + 15 

= 36 units 

 So the answer is 36 units
I hope my answer helped you.

IRINA_888 [86]3 years ago

4 0

Answer:

36units

Step-by-step explanation:

yeah thatss right

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In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

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Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
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Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$