Another identity problem...

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

3 0

Answer:

cot(x)=-sqrt(3)/3

Step-by-step explanation:

I'm going to use the Pythagorean Identity sin^2(x)+cos^2(x)=1 to find the sine value since cotangent value is the cosine value over sine value.

Plug in:

sin^2(x)+(-1/2)^2=1

Simplify:

sin^2(x)+1/4=1

Subtract 1/4 on both sidea:

sin^2(x)=3/4

Square root both sides:

sin(x)=sqrt(3)/2. [Sine is positive since x is in 2nd quadrant]

cot(x)=(-1/2)/(sqrt(3)/2))

cot(x)=-1/sqrt(3)

Rationalize denominator by multiplying by sqrt(3)/sqrt(3):

cot(x)=-sqrt(3)/3

cot(x)=

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The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

ad-work [718]

Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.