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if i am given two sides of a triangle with the angle in between how do i find the third angle

1. Example: Given b,c and   $$\angle CAB$$  (A) between:

$$\tan{(B)} = \dfrac{ b\cdot \sin{(A)} }{ c-b\cdot \cos{(A)}}\qquad
\tan{(C)} = \dfrac{ c\cdot \sin{(A)} }{ b-c\cdot \cos{(A)}}$$

cycle:

2. Example: Given c,a and   $$\angle ABC$$  (B) between:

$$\tan{(C)} = \dfrac{ c\cdot \sin{(B)} }{ a-c\cdot \cos{(B)}}\qquad
\tan{(A)} = \dfrac{ a\cdot \sin{(B)} }{ c-a\cdot \cos{(B)}}$$

cycle:

3. Example: Given a,b and   $$\angle BCA$$  (C) between:

$$\tan{(A)} = \dfrac{ a\cdot \sin{(C)} }{ b-a\cdot \cos{(C)}}\qquad
\tan{(B)} = \dfrac{ b\cdot \sin{(C)} }{ a-b\cdot \cos{(C)}}$$

If we use (x+) to indicate the following sum
   1 + 2 + 3 + ... + x
then find the value of k in the following equation
(28+) - (27+) = (k+)

$$\small{\text{$
\begin{array}{rcl}
(28+)-(27+) = (k+) = 28 &=& S_k \qquad (28+) = S_{28} \quad (27+)=S_{27}\\\\
S_{28}-S_{27} = t_{28} = 28&=&S_k\\\\
S_k &=& t_1 \cdot \dbinom{k}{1} + d\cdot \dbinom{k}{2} \qquad t_1=d=1\\\\
S_k &=& \dbinom{k}{1} + \dbinom{k}{2} = \dbinom{k+1}{2} = 28\\\\ \\
\dfrac{k(k+1)}{2} &=& 28 \\\\
k(k+1) &=& 56\\\\
k^2+k-56&=&0\\\\
\Rightarrow k &=& \dfrac{-1+15}{2} = 7\\\\
\mathbf{S_k = (k+) }& \mathbf{=} & \mathbf{(7+)}\\\\
(7+)=1+2+3+4+5+6+7&=&28 \\
\end{array}
$}}$$

1+2+3+4+5+6+7+8+9+10 = 55  What is the Fastest way to do that?

$$\small{\text{$
1+2+3+4+5+6+7+8+9+10 =
\dbinom{n+1}{2} = \dbinom{10+1}{2} = \dbinom{11}{2}
= \dbinom{11}{9}=\dfrac{10\cdot 11}{2} = 5\cdot 11 = 55
$}}$$

$$\small{\text{$
\begin{array}{lrcl}
1. & \sin{(A)} &=& -\frac 35 \quad \Rightarrow \quad A = \arcsin{(-\frac 35 )} = -36.87\ensurement{^{\circ}} ~ (IV.)\\\\
2. & \sin{(180\ensurement{^{\circ}}-A)} &=& -\frac 35 \quad \Rightarrow \quad A = 180\ensurement{^{\circ}}- \arcsin{(-\frac 35 )} = 180\ensurement{^{\circ}} + 36.87\ensurement{^{\circ}} = 216.87\ensurement{^{\circ}} ~(III.)\\\\
\end{array}
$}}$$

And each sqrt has a plus and a minus

In terms of electromagnetism, one watt is the rate at which work is done when one ampere (A) of current flows through an electrical potential difference of one volt (V).

And each rate of power is to multiply  with the hour you use it.

What is cos[sin^-1(-3/5)]

$$\boxed{~\cos{[\arcsin{(x)}]} = \pm \sqrt{1-x^2}~}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{\left[\arcsin{ \left( -\dfrac 35 \right) } \right]} &=&
\pm \sqrt{1-\left(-\dfrac 35 \right)^2}\\\\
\cos{\left[\arcsin{ \left( -\dfrac 35 \right) } \right]} &=&
\pm \sqrt{1- \dfrac{9}{25} }\\\\
\cos{\left[\arcsin{ \left( -\dfrac 35 \right) } \right]} &=&
\pm \sqrt{\dfrac{16}{25} }\\\\
\mathbf{ \cos{\left[\arcsin{ \left( -\dfrac 35 \right) } \right]} } & \mahtbf{=} &
\mathbf{ \pm \dfrac{4}{5} }\\\\
\end{array}
$}}$$

The second term of a geometric sequence is 2 and the last term is 8. If the common ratio is 2√, how many geometric means are there between the first term and the last term?

$$\small{\text{$
\begin{array}{rcl}
t_2 &=& t_1\cdot r^1 \qquad | \qquad t_2 = 2 \quad \text{ and }\quad r = 2^{\frac 12} \\
2 &=& t_1 \cdot 2^{\frac 12}\\
t_1 &=& 2\cdot 2^{-\frac 12}\\
t_1 &=& 2^{ 1- \frac 12 }\\
\mathbf{t_1} & \mathbf{=} & \mathbf{2^{ \frac 12 }}\\\\\\
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
t_n &=& t_1\cdot r^{n-1} \qquad | \qquad t_n = 2^3 \quad \text{ and }\quad t_1 = 2^{ \frac 12 } \quad \text{ and } \quad r = 2^{\frac 12}\\
2^3 &=& 2^{ \frac 12 } \cdot 2^{\frac 12 \cdot(n-1)} \qquad | \qquad 1^2\\
2^6 &=& 2^1 \cdot 2^{n-1}\\
2^6 &=& 2^{1+n-1}\\
2^6 &=& 2^n\\
6 &=& n\\
\mathbf{ n } & \mathbf{=} & \mathbf{6} \\
\end{array}
$}}$$

We have 6 terms, so between the first term and the 6th term are 4 terms.

$$\small{\text{$
\begin{array}{rcl}
t_1 &=& 2^{\frac 12} = \sqrt{2}\\
t_2 &=& 2^{\frac 22} = 2 \\
t_3 &=& 2^{\frac 32} = 2^{\frac 22}\cdot 2^{\frac 12} = 2\cdot \sqrt{2}\\
t_4 &=& 2^{\frac 42} = 2^2 = 4\\
t_5 &=& 2^{\frac 52} = 2^{\frac 42}\cdot 2^{\frac 12} = 2^2\cdot \sqrt{2}= 4\cdot \sqrt{2}\\
t_6 &=& 2^{\frac 62} = 2^3 = 8\\
\end{array}
$}}$$

A vacuum cleaner has a power rating of  600W

For how many hours was the vacuum cleaner used if the cost of operating the vacuum cleaner was $6.36?

The cost of electricity is $0.2650 per kilowatt-hour.

600 W = 0.6 kW

$$\small{\text{$
\begin{array}{rcl}
0.6~ \mathrm{kW}\cdot t
\cdot \left(
\dfrac{\$0.2650}{\mathrm{kW\cdot h}}
\right) &=& \$6.36 \\\\
0.6\cdot t
\cdot \left(
\dfrac{0.2650}{\mathrm{h}}
\right) &=& 6.36 \\\\
t &=& \dfrac{ 6.36 }{0.2650\cdot0.6}~ \mathrm{h} \\\\
\mathbf{t} & \mathbf{=} & \mathbf{40~\mathrm{h}} \\\\
\end{array}
$}}$$

Check:

$$\small{\text{$
\begin{array}{rcl}
0.6~ \mathrm{kW}\cdot 40~ \mathrm{h} &=& 24 ~\mathrm{kW~ h}\\\\
24 ~\mathrm{kW~ h} \cdot \dfrac { \$0.2650 } {\mathrm{kW~ h} } &=& \$6.36
\end{array}
$}}$$

Equivalent Ratios 2:6=   :18=10:    =    :72=200:

$$\small{\text{$
\begin{array}{rclcl}
&&\dfrac{\textcolor[rgb]{0,0,1}{2}}{\textcolor[rgb]{0,0,1}{6}} &=& \dfrac{1}{3}\\\\
&=&\dfrac{1}{3}\cdot \dfrac{6}{6} &=& \dfrac{\textcolor[rgb]{1,0,0}{6}}{\textcolor[rgb]{0,0,1}{18}}\\\\
&=&\dfrac{1}{3}\cdot \dfrac{10}{10} &=& \dfrac{\textcolor[rgb]{0,0,1}{10}}{\textcolor[rgb]{1,0,0}{ 30}}\\\\
&=&\dfrac{1}{3}\cdot \dfrac{24}{24}&=& \dfrac{\textcolor[rgb]{1,0,0}{ 24}}{\textcolor[rgb]{0,0,1}{72}}\\\\
&=&\dfrac{1}{3}\cdot \dfrac{200}{200} &=&\dfrac{\textcolor[rgb]{0,0,1}{200}}{ \textcolor[rgb]{1,0,0}{ 600}}
\end{array}
$}}$$

2:6 = 6:18 = 10:30 = 24:72 = 200:600

Find all values of  such that .

$$\small{\text{$
\begin{array}{rcl}
\sqrt{4x^2}-\sqrt{x^2} &=& 6\\
\sqrt{4}\sqrt{x^2}-\sqrt{x^2} &=& 6\\
2\sqrt{x^2}-\sqrt{x^2} &=& 6\\
\sqrt{x^2} &=& 6 \qquad | \qquad 1^2\\
x^2 &=& 36 \qquad | \qquad \pm\sqrt{}\\
x_{1,2} &=& \pm\sqrt{36} \\
x_{1,2} &=& \pm6 \\
\mathbf{x_1} & \mathbf{=} & \mathbf{6}\\
\mathbf{x_2} & \mathbf{=} & \mathbf{-6}
\end{array}
$}}$$