All Answers 356020 Answers

how do I get angle of a triangle not right angled when I know three sides

\(Cosine \ set\)

\(a^2=b^2+c^2-2bc\ cos\alpha\)

\(\alpha=arccos\frac{b^2+c^2-a^2}{2bc}\)

\(\beta=arccos\frac{a^2+c^2-b^2}{2ac}\)

\(\gamma=arccos\frac{a^2+b^2-c^2}{2ab}\)

  !

Not sure of the utility of your question, but

58 degrees x 16.2 = 939.6 degrees

Sounds alright 

x = number of eggs originally

x - (4/9x +2) = 28

5/9x -2=28

5/9x = 30

x = 30 * 9/5 = 54 eggs

1.204* 10²⁴

^-mathgeek

Good question it is...

3/8=  0.375

-mathgeek

5/-5= -1-7=-8

-mathgeek

Tysm fr the answer!!😊😊

To be jokeing yes 2+2= can be 22. But if you use math it accualy is 2+2=4

-mathgeek

84/3 = 28

Ty for the answer. Could ya solve the others as soon as possible?

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)

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

21 is the answer

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:

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

Be = 9.0121 gm/mole

Cl= 35.45

Therefore   Be Cl2 = 79.9121 gm / formula unit

179 g / 79.9121 = 2.24 formula units

.075 x mole = .075 x Avagadro's Number

= .075 x 6.022 x 10^23 = 4.517 x 10^22   ions

18 n*gga

What does PIE equal?

\(\pi = 4\cdot \arctan(1) \qquad \text{angle in rad}\)

Need to be careful here heureka

\(\sqrt{-2}\rightarrow i\sqrt2\\ \sqrt{-12}\rightarrow 2i\sqrt3\)

Multiply these together and you should get a negative result.

....And also , I think

   sqrt(-2) x sqrt(-12) = i sqrt 2 x i sqrt 12  = i^2 sqrt(24) = i^2 (2) sqrt 6 = -2 sq rt 6