$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill$

$\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}$

now that we know how long each one is, let's plug those in Heron's Area formula.

$\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18$

5 0

3 years ago

Find m∠A given ΔABC where a=4, b=6, c=3.

igomit [66]

For this question, we're going to use the law of cosines.
The law of cosines is the following equation:

$a^2=b^2+c^2-2(b)(c) \cos(A)$

We know the values of a, b, and c. We want to find the measure of angle A.
$a=4$
$b=6$
$c=3$

Now, plug in these values into the equation.

$4^2=6^2+3^2-2(3)(6) \cos(A)$
$16=45-36 \cos(A)$

Add both sides by $36 \cos(A)$ and subtract both sides by 16

$36 \cos(A)=29$

Divide both sides by 36

$\cos(A)= \dfrac{29}{36}$

Take the inverse cosine, or arc cosine, of both sides.
Using a calculator, you'll get the following result.

$A \approx 36.3^O$

The measure of angle A should be 36.3 degrees. Hope this helps! :)

4 0

3 years ago

PLEASE I NEED HELP I HAVE NO IDEA WHAT THE ANSWERS ARE

Ivenika [448]

Answer:

Sector AOB: 3.87

Sector BOC: 7.74

Sector COD: 5.16

Sector DOE: 3.01

Sector AOE: 11.19

Explanation:

Equation used: $S=\frac{r^2*a}{2}$

4 0

2 years ago