heureka

580 nm light shines on a double slit with d=0.000125 m.
What is the angle of the third dark interference minimum(m=3)?

\(\begin{array}{rcll} Let\ d &=& 0.000125\ m \\ d &=& 125 \cdot 10^{-6}\ m \\\\ Let\ \lambda &=& 580\ nm \\ \lambda &=& 580 \cdot 10^{-9}\ m \\ \end{array} \)

\(\begin{array}{|rcll|} \hline d\sin(\theta) &=& m \lambda \\ \sin(\theta) &=& m\cdot \frac{\lambda}{d} \quad & | \quad m=3 \\ \sin(\theta) &=& 3\cdot \frac{580 \cdot 10^{-9}\ m }{125 \cdot 10^{-6}\ m} \\ \sin(\theta) &=& 3\cdot \frac{580}{125} \cdot 10^{-9+6} \\ \sin(\theta) &=& \frac{1740}{125} \cdot 10^{-3} \\ \sin(\theta) &=& 13.92 \cdot 10^{-3} \\ \sin(\theta) &=& 0.01392 \\ \theta &=& \arcsin(0.01392) \\ \theta &=& 0.79758300970^{\circ} \\ \hline \end{array} \)

if x+ 1/x= 2, what is x2 + 1/x2

if  \(x+ \frac{1}{x}= 2\) , what is \( x^2 + \frac{1}{x^2}\)

\(\begin{array}{|rcll|} \hline x+ \frac{1}{x} &=& 2 \quad & | \quad \text{square both sides} \\ \left(~x+ \frac{1}{x}~\right)^2 &=& 2^2 \\ \left(~x+ \frac{1}{x}~\right)^2 &=& 4 \\ x^2 +2\cdot x\cdot \frac{1}{x} + \frac{1}{x^2} &=& 4 \\ x^2 +2 + \frac{1}{x^2} &=& 4 \quad & | \quad -2 \\ x^2 + \frac{1}{x^2} &=& 4-2 \\ \mathbf{x^2 + \frac{1}{x^2}} & \mathbf{=} & \mathbf{2} \\ \hline \end{array} \)

At what point do the graph of the equations shown below intersect?

-8x+2y=4

6x+3y=-9

Intersection at  point (-0.83333333333,-1.33333333333) or (\( -\frac{5}{6},-\frac{4}{3}\) )

Cotangent of pi/5 in decimals.

\(\begin{array}{|rcll|} \hline && \cot(\frac{\pi}{5}) \\ &=& \cot( 36^{\circ} ) \\ &=& \frac{ 1 } { \tan(36^{\circ}) } \\ &=& \frac{ 1 } { \sqrt{5-2\cdot \sqrt{5} } } \\ &=& 1.3763819204711735382 \\ \hline \end{array}\)

Line AB, CB, and DA are the same lengths. Angles 1 and 2 are the same.The angle of A is 120o .

The area of polygon ABCD is 60.

What is the area of the triangle ABD?

Let angles 1 and 2 are \(\varphi\)

Let AB, CB, and DA = a

1. \(DB =\ ?\)

\(\begin{array}{|rcll|} \hline DB^2 &=& a^2+a^2-2\cdot a\cdot a\cdot \cos(120^{\circ}) \\ DB^2 &=& 2a^2[~1-\cos(120^{\circ}) ~] \quad & | \quad \cos(120^{\circ}) = \cos(180^{\circ}-60^{\circ}) = -\cos(60^{\circ}) \\ DB^2 &=& 2a^2[~1+\cos(60^{\circ}) ~] \quad & | \quad \cos(60^{\circ}) = \frac12 \\ DB^2 &=& 2a^2(~1+\frac12 ~) \\ DB^2 &=& 2a^2(~\frac32 ~) \\ DB^2 &=& 3a^2 \\ \mathbf{DB} & \mathbf{=} & \mathbf{a\cdot \sqrt{3}} \\ \hline \end{array}\)

2. \(\varphi =\ ?\)

\(\begin{array}{|rcll|} \hline \frac{ \sin(\varphi) } {a} &=& \frac{\sin(120^{\circ})} {DB} \\ \frac{ \sin(\varphi) } {a} &=& \frac{\sin(120^{\circ})} {a\cdot \sqrt{3}} \\ \sin(\varphi) &=& \frac{\sin(120^{\circ})} {\sqrt{3}} \quad & | \quad \sin(120^{\circ}) = \sin(180^{\circ}-60^{\circ}) = \sin(60^{\circ}) \\ \sin(\varphi) &=& \frac{\sin(60^{\circ})} {\sqrt{3}} \quad & | \quad \sin(60^{\circ}) = \frac{\sqrt{3}} {2} \\ \sin(\varphi) &=& \frac{\frac{\sqrt{3}} {2} } {\sqrt{3}} \\ \sin(\varphi) &=& \frac12\\ \mathbf{\varphi} & \mathbf{=} & \mathbf{30^{\circ}} \\ \hline \end{array} \)

3. Angle \(C =\ ?\)

\(\begin{array}{|rcll|} \hline \frac{ \sin(C) } {DB} &=& \frac{ \sin{30^{\circ}} } {a} \quad & | \quad \sin(30^{\circ}) = \frac12 \\ \frac{ \sin(C) } {a\cdot \sqrt{3}} &=& \frac{ \frac12 } {a} \\ \sin(C) &=& \frac{ \sqrt{3} } {2} \\ \mathbf{C} & \mathbf{=} & \mathbf{60^{\circ}} \\ \hline \end{array}\)

So angle CBD is \(90^{\circ}\) and triangle CBD is a right angular triangle.

4. The area of triangle CBD:

\(\begin{array}{|rcll|} \hline A_{CBD} &=& \frac{ a\cdot DB }{2} \\ &=& \frac{ a\cdot a\cdot \sqrt{3} }{2} \\ \mathbf{A_{CBD}} & \mathbf{=} & \mathbf{\frac{ a^2\sqrt{3} }{2}} \\ \hline \end{array} \)

5. The area of triangle ABD:

\(\begin{array}{|rcll|} \hline A_{ABD} &=& \frac{ DB \cdot a \cdot \sin(\varphi) }{2} \\ &=& \frac{ a\cdot \sqrt{3} \cdot a \cdot \frac12 }{2} \\ A_{ABD} &=& \frac{ a^2\cdot \sqrt{3} }{4} \\ \text{or}\\ a^2 &=& \frac{4\cdot A_{ABD} } {\sqrt{3}} \\ \mathbf{a^2} & \mathbf{=} & \mathbf{\frac{4\cdot A_{ABD} } {\sqrt{3}}} \\ \hline \end{array} \)

The area of polygon ABCD \(A_{ABCD} = 60\)

\(\begin{array}{|rcll|} \hline A_{ABCD} &=& A_{ABD} + A_{CBD} \quad & | \quad A_{CBD} = \frac{ a^2\sqrt{3} }{2} \qquad A_{ABCD} = 60 \\ 60 &=& A_{ABD} + \frac{ a^2\sqrt{3} }{2} \quad & | \quad a^2 = \frac{4\cdot A_{ABD} } {\sqrt{3}} \\ 60 &=& A_{ABD} + \frac{ \frac{4\cdot A_{ABD} } {\sqrt{3}}\sqrt{3} }{2} \\ 60 &=& A_{ABD} + 2\cdot A_{ABD} \\ 60 &=& 3\cdot A_{ABD} \\ 20 &=& A_{ABD} \\ \mathbf{A_{ABD}} & \mathbf{=} & \mathbf{20} \\ \hline \end{array}\)

The area of the triangle ABD is 20

What is the largest integer $n$ such that $7^n$ divides $1000!$ ?

1. We calculate 1000 in base 7:

\(\begin{array}{rcll} 1000_{10} &=& 2626_7 \\ \end{array} \)

2. The sum of the digits in base 7 is: \(\begin{array}{rcll} 2+6+2+6 = 16 \end{array} \)

3. \(n =\ ?\)

\(\begin{array}{|rcll|} \hline n &=&\frac{1000-(\text{sum of the standard base-p digits of 1000})}{7-1} \\ n &=&\frac{1000-16}{6} \\ n &=&164 \\ \hline \end{array} \)

\(7^{164} \) will be divide 1000! and \(7^{164} \) is one prime factor of 1000! for prime = 7.\(\)

What is the largest integer $n$ such that $7^n$ divides $1000!$ ?

In mathematics, Legendre's formula gives an expression for the exponent of the largest power of a prime p that divides the factorial  \({\displaystyle n!} \)

Applications to Trignometry

1.The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 m towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower. 

Let x the height of the tower

Let y the distance to the tower, if the observer moves 20m towards the tower.

\(\begin{array}{|rcll|} \hline \tan(45^{\circ}) = \frac{x}{y} &=& 1 \qquad & | \qquad \tan(45^{\circ}) = 1 \\ y &=& x \\\\ \tan(30^{\circ}) = \frac{x}{20+y} &=& \frac{1}{\sqrt{3}} \qquad & | \qquad \tan(30^{\circ}) = \frac{1}{\sqrt{3}} \\ \frac{x}{20+y} &=& \frac{1}{\sqrt{3}} \qquad & | \qquad y = x \\\\ \frac{x}{20+x} &=& \frac{1}{\sqrt{3}} \\ \dots \\ x &=& \frac{ 20 } { \sqrt{3} -1 } \\ \mathbf{x} & \mathbf{=} & \mathbf{27.32\ m} \\ \hline \end{array}\)

4. The angle of elevation of the Top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y 40m vertically above X the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX(use root 3= 1.73)

Let x = PQ (height )
Let y = PX (distance)

\(\begin{array}{|rcll|} \hline \tan(60^{\circ}) = \frac{x}{y} &=& \sqrt{3} \qquad & | \qquad \tan(60^{\circ}) = \sqrt{3} \\ y &=& \frac{x}{\sqrt{3}} \\\\ \tan(45^{\circ}) = \frac{x-40}{y} &=& 1 \qquad & | \qquad \tan(45^{\circ}) = 1 \\ x-40 &=& y \\\\ x-40 &=& \frac{x}{\sqrt{3}} \\ \dots \\ x &=& \frac{ 40\cdot \sqrt{3} } {\sqrt{3}-1} \qquad & | \qquad \sqrt{3} = 1.73 \\ &=& \frac{ 40\cdot 1.73 } {1.73-1} \\ &=& \frac{ 69.2 } {0.73} \\ \mathbf{x} & \mathbf{=} & \mathbf{94.79\ m} \\\\ y &=& \frac{x}{\sqrt{3}} \\ &=& \frac{94.79}{\sqrt{3}} \qquad & | \qquad \sqrt{3} = 1.73 \\ &=& \frac{94.79}{1.73} \\ \mathbf{y} & \mathbf{=} & \mathbf{54.79\ m} \\\\ \hline \end{array} \)

please answer and simplify squareroot of -2 times square root of -12. Remember i factor.

\(\sqrt{-2}\cdot \sqrt{-12} =\ ?\)

\(\sqrt{z} \qquad z \in \mathbb{Q}\) has two solutions in the complex plane.

1. Wolfram Alpha calculate for \(\sqrt{-2}\):

2. Wolfram Alpha calculate for \(\sqrt{-12}\):

So the product is:

\(\begin{array}{|rcll|} \hline && \sqrt{-2}\cdot \sqrt{-12} \\ &=& (\pm i\cdot\sqrt{2})\times (\pm 2i\sqrt{3}) \\ \hline \end{array} \)

Here we have four solutions:

\(\begin{array}{|rcll|} \hline 1. & (+ i\cdot\sqrt{2})\times(+ 2i\sqrt{3}) &=& -2\sqrt{6} \\ 2. & (+ i\cdot\sqrt{2})\times(- 2i\sqrt{3}) &=& +2\sqrt{6} \\ 3. & (- i\cdot\sqrt{2})\times(+ 2i\sqrt{3}) &=& +2\sqrt{6} \\ 4. & (- i\cdot\sqrt{2})\times(- 2i\sqrt{3}) &=& -2\sqrt{6} \\ \hline \end{array}\)

and finally:

\(\begin{array}{|rcll|} \hline \sqrt{-2}\cdot \sqrt{-12} = \pm 2\sqrt{6} \\ \hline \end{array}\)