heureka

7)
In triangle ABC, sin A = 4/5. Find cos B.
Picture: https://latex.artofproblemsolving.com/c/8/3/c83240e209d830cd2275ed9f771185a1d1c17733.png

sin-rule

\(\begin{array}{|rcll|} \hline \dfrac{\sin(B)} {10} &=& \dfrac{\sin(A)}{17} \\\\ \sin(B) &=& \dfrac{10\cdot \sin(A)}{17} \quad & | \quad \sin(A) = \dfrac {4}{5} \\\\ \sin(B) &=& \dfrac{10\cdot \dfrac {4}{5}}{17} \\\\ \sin(B) &=& \dfrac{2\cdot 4 }{17} \\\\ \mathbf{\sin(B)} & \mathbf{=} & \mathbf{ \dfrac{8}{17} } \\\\ \cos(B) &=& \sqrt{1-\sin^2(B)} \\ \\ \cos(B) &=& \sqrt{1- \left(\dfrac{8}{17} \right)^2 } \\ \\ \cos(B) &=& \sqrt{1- \dfrac{8^2}{17^2} } \\ \\ \cos(B) &=& \dfrac{ \sqrt{17^2-8^2} } {17} \\ \\ \cos(B) &=& \dfrac{ \sqrt{225} } {17} \\ \\ \mathbf{\cos(B)} & \mathbf{=} & \mathbf{ \dfrac{ 15 } {17} } \\ \hline \end{array}\)

6)

Find cos A.
Picture: https://latex.artofproblemsolving.com/6/4/a/64ae47768bae6b24ab792c44a92c5de244955cad.png

cos-rule

\(\begin{array}{|rcll|} \hline 2^2 &=& 3^2+4^2-2\cdot 3 \cdot 4\cdot \cos(A) \\ 4 &=& 9+ 16 -24\cdot \cos(A) \\ 4 &=& 25 -24\cdot \cos(A) \quad & | \quad +24\cdot \cos(A) \\ 4 +24\cdot \cos(A) &=& 25 \quad & | \quad -4 \\ 24\cdot \cos(A) &=& 25 -4 \\ 24\cdot \cos(A) &=& 21 \quad & | \quad : 24 \\ \cos(A) &=& \dfrac{21} {24} \\\\ \mathbf{\cos(A)} & \mathbf{=} & \mathbf{ \dfrac{7} {8} } \\ \hline \end{array}\)

5)
Find sin B.
Picture:https://latex.artofproblemsolving.com/4/7/a/47a312412eff41c09e4ce21e5edbb7e52e8b151c.png

\(\begin{array}{|rcll|} \hline AD^2 + 6^2 &=& 10^2 \\ AD^2 & = & 10^2 - 6^2 \\ AD^2 & = & 64 \\ \mathbf{AD} & \mathbf{=} & \mathbf{ 8 } \\\\ AB^2 &=& 15^2 + AD^2 \quad & | \quad AD^2 = 64 \\ AB^2 &=& 15^2 + 64 \\ AB^2 &=& 289 \\ \mathbf{AB} & \mathbf{=} & \mathbf{17 } \\\\ \sin(B) &=& \dfrac {AD} {AB} \\\\ \mathbf{\sin(B)} & \mathbf{=} & \mathbf{ \dfrac {8} {17}} \\ \hline \end{array}\)

3)
In triangle ABC, CA = 4sqrt2, CB = 4sqrt3, and angle A = 60 degrees. What is B in degrees?

sin-rule

\(\begin{array}{|rcll|} \hline \dfrac{\sin(B)} {4\sqrt{2}} &=& \dfrac{\sin(60^{\circ})} {4\sqrt{3}} \\\\ \dfrac{\sin(B)} { \sqrt{2}} &=& \dfrac{\sin(60^{\circ})} { \sqrt{3}} \quad & | \quad \sin(60^{\circ}) = \frac{ \sqrt{3} } {2} \\\\ \dfrac{\sin(B)} {\sqrt{2}} &=& \dfrac{ \sqrt{3} } {2\sqrt{3}} \\\\ \dfrac{\sin(B)} {\sqrt{2}} &=& \dfrac{ 1 } {2 } \\\\ \sin(B) &=& \dfrac{ \sqrt{2} } {2 } \\\\ \mathbf{B} & \mathbf{=} & \mathbf{45^{\circ}} \\ \hline \end{array}\)

2)

Find AC. 

Picture: https://latex.artofproblemsolving.com/6/8/d/68dd918aac5d85a0b2a5b3b9c6a79eabd3669f01.png

sin-rule

\(\begin{array}{|rcll|} \hline \dfrac{AC}{\sin(45^{\circ})} &=& \dfrac{30} { \sin\Big(180^{\circ}-(75^{\circ}+45^{\circ}) \Big) } \quad & | \quad \small{\sin\Big(180^{\circ}-(75^{\circ}+45^{\circ}) \Big) = \sin(75^{\circ}+45^{\circ})} \\\\ \dfrac{AC}{\sin(45^{\circ})} &=& \dfrac{30} { \sin(75^{\circ}+45^{\circ}) } \\\\ \dfrac{AC}{\sin(45^{\circ})} &=& \dfrac{30} { \sin(120^{\circ}) } \\\\ AC &=& \dfrac{30\cdot \sin(45^{\circ})} { \sin(120^{\circ}) } \quad & | \quad \sin(45^{\circ}) = \frac{\sqrt{2}} {2} \\\\ AC &=& \dfrac{30\cdot \frac{\sqrt{2}} {2}} { \sin(120^{\circ}) } \quad & | \quad \sin(120^{\circ} = \frac{\sqrt{3}} {2} \\\\ AC &=& \dfrac{30\cdot \frac{\sqrt{2}} {2}} { \frac{\sqrt{3}} {2} } \\\\ AC &=& \dfrac{30\cdot \sqrt{2} } { \sqrt{3} } \cdot \frac{ \sqrt{3} } {\sqrt{3}} \\\\ AC &=& \dfrac{30\cdot \sqrt{2} \sqrt{3} } { 3 } \\\\ AC &=& 10\cdot \sqrt{6} \\ AC &=& 24.4948974278 \\ \hline \end{array}\)

1)

Find BC.

Picture: https://latex.artofproblemsolving.com/f/e/4/fe45662f573f87e30f9f30d0b1ef1f94b4c22fc2.png

cos-rule:

\(\begin{array}{|rcll|} \hline BC^2 &=& 6^2 +(7\sqrt{2})^2-2\cdot 6 \cdot7\sqrt{2}\cdot \cos(45^{\circ}) \quad & | \quad \cos(45^{\circ}) = \frac{\sqrt{2}}{2} \\ BC^2 &=& 36 + 49\cdot 2 -2\cdot 6 \cdot7\sqrt{2}\cdot \frac{\sqrt{2}}{2} \\ BC^2 &=& 36 + 98 -2\cdot 6 \cdot7 \cdot \frac{2}{2} \\ BC^2 &=& 36 + 98 -2\cdot 6 \cdot7 \\ BC^2 &=& 50 \\ BC &=&\sqrt{ 50} \\ BC &=&\sqrt{ 5^2\cdot 2} \\ BC &=& 5\sqrt{2} \\ \hline \end{array}\)

How do you solve
$i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}$

\(\small{ \begin{array}{|rcll|} \hline && i^{-100}+i^{-99}+i^{-98}+i^{-97}+\cdots+i^{-2}+i^{-1}+i^0+i^1+i^2+i^3+i^4\cdots+i^{99}+i^{100} \\ &=& i^{-100}(1+i)+i^{-98}(1+i) +\cdots+i^{-2}(1+i)+i^0+i^1(1+i)+i^3(1+i)+\cdots+i^{99}(1+i) \\ && \boxed{\color{red}1+i =1-1=0} \\ &=& i^{-100}\times 0+i^{-98}\times 0 +\cdots+i^{-2}\times 0+i^0+i^1\times 0+i^3\times 0+\cdots+i^{99}\times 0 \\ &=&0+i^0+0 \\ &=& i^0 \\ &\mathbf{=}& \mathbf{1} \\ \hline \end{array} }\)

​Classify by degree and number of terms.

The points $B(1, 1)$, $I(2, 4)$ and $G(5, 1)$ are plotted in the standard rectangular coordinate system to form triangle $BIG$.
Triangle $BIG$ is translated five units to the left and two units upward to triangle $BIG'$,
in such a way that
$B'$ is the image of $B$,
$I'$ is the image of $I$, and
$G'$ is the image of $G$.
What is the midpoint of segment $B'G'$? Express your answer as an ordered pair.

1.

\(\text{The midpoint of $BG$ is $\frac{B+G}{2}$ }\)

\(\begin{array}{|rcll|} \hline && \dfrac{B+G}{2} \\\\ &=& \dfrac{\dbinom{1}{1}+\dbinom{5}{1}}{2} \\\\ &=& \dfrac{\dbinom{1+5}{1+1}}{2} \\\\ &=& \dfrac{\dbinom{6}{2}}{2} \\\\ &=& \dbinom{\frac{6}{2}}{\frac{2}{2}}\\\\ &=& \dbinom{3}{1}\\ \hline \end{array}\)

The midpoint of BG is \(\mathbf{(3,\ 1)}\)

2.

\(\text{The translation of the midpoint of $BG$ is the midpoint of $B'G'$ }\)

\(\begin{array}{|ll|} \hline \text{Translation}\\ x' = x-5 \quad & \text{five units to the left} \\ y' = y+2 \quad & \text{two units upward} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \underbrace{\dbinom{3}{1}}_{\text{midpoint of $BG$ }} \rightarrow \quad \underbrace{\dbinom{3-5}{1+2}}_{\text{translation}} &=& \underbrace{ \mathbf{ \dbinom{-2}{3}} }_{ \text{midpoint of $B'G'$} } \\ \hline \end{array}\)

The midpoint of B'G' is \(\mathbf{(-2,\ 3)}\)

A senate committee has 5 Republicans and 4 Democrats. In how many ways can the committee members sit in a row of 9 chairs, such that all 4 Democrats sit together?

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