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This question has been asked and answered about a billion times.

I did this:

= 60^2+74^2-2(60)(74)cos(85)

=3600+5476-8880*0.087155743

=8302.057002

sqrt of 8302.057002  = 91.11 miles

Is this right?

And thank you for your answer! It helped me understand it better

But of course! I had to subtract 38-32 to get 6. I see it now. Thank you so much! your answers are so easy to understand

A surveyor determines that the angle of elevation to the top of a vertical building from a pt on level ground is 38deg.

She then moves back 50 feet & determines that the angle of elevation is 32 deg.

Find the height of the building.

\(\begin{array}{|rcll|} \hline \dfrac{\sin(6^\circ)}{50} &=& \dfrac{\sin(32^\circ)}{y} \\ \mathbf{y} &=& \mathbf{ \dfrac{50\sin(32^\circ)}{\sin(6^\circ)} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \sin(38^\circ) &=& \dfrac{h}{y} \\ h &=& y\sin(38^\circ) \\ h &=& \dfrac{50\sin(32^\circ)\sin(38^\circ)}{\sin(6^\circ)} \\ \mathbf{h} &=& \mathbf{ 156\ \text{feet} } \\ \hline \end{array} \)

It could equally well be - infinity. 

So I would call it undefined.

For starters you have not labeled  the sides of your triangle properly.

Side a  must be  opposite angle A

Same for b and c

Always little anglers for sides and the Capital equivalent for the opposite angle.

After that you will need to use sine  rule to get another angles

and then angle sum of a triangle to get the third one.

You can display what you do for each step if you want. People will give you feedback.   

You have uses Pythagoras's theorem.

This only works if the angle is 90 degrees,  Your angle is 48+37

You have to use the cosine rule to solve this.

Here is a tutorial on it if you need one.

If cos 3x + cos x = 0, find all possible values of cos x.

Hello Guest!

\({\color{blue}cos( 3x)=4cos^3( x)-3cos( x)}=(4cos^2(x)-3)\cdot cos(x)\)  \(Bartsch\ Leipzig\ 1980\)

\(cos 3x + cos x = 0\)

\((4cos^2(x)-3)\cdot cos(x)+cos(x)=0\)

\(\large cos(x)=u\)

\((4u^2-3)\cdot u+u=0\\ 4u^3-3u+u=0\\ u(4u^2-3+1)=0\)

\(u_1=0\)

\(4u^2-2=0\\ u^2=\frac{1}{2}\)

\(u_2=+\sqrt{\frac{1}{2}}=\frac{1}{2}\sqrt{2}\\ u_3=-\sqrt{\frac{1}{2}}=-\frac{1}{2}\sqrt{2}\)

\(cos(x_1)=0\\ cos(x_2)=\frac{1}{2}\sqrt{2}\\ cos(x_3)=-\frac{1}{2}\sqrt{2}\)

  !

Find

\(\sin\left( \arcsin\left(\dfrac{3}{5}\right) + \arccos\left(\dfrac{15}{17}\right) \right)\).

\(\text{Let $x=\dfrac{3}{5},\ y=\dfrac{15}{17} $}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\sin\left( \arcsin\left(\dfrac{3}{5}\right) + \arccos\left(\dfrac{15}{17}\right) \right)} \\\\ &=& \sin\Big( \underbrace{\arcsin\left(x\right)}_{\alpha} + \underbrace{\arccos\left(y\right)}_{\beta} \Big) \\ && \boxed{\text{Formula: } \sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+ \cos(\alpha)\sin(\beta)} \\\\ &=& \sin\Big(\arcsin\left(x\right)\Big)\cos\Big(\arccos\left(y\right)\Big) + \cos\Big(\arcsin\left(x\right)\Big)\sin\Big(\arccos\left(y\right)\Big) \\ &=& xy + \mathbf{\cos\Big(\arcsin\left(x\right)\Big)}\sin\Big(\arccos\left(y\right)\Big) \\ && \boxed{ \arcsin\left(x\right) = \alpha\\ x=\sin(\alpha)\\ x^2=\sin^2(\alpha)\\ x^2=1-\cos^2(\alpha)\\ \cos^2(\alpha)=1-x^2 \\ \cos(\alpha) =\sqrt{1-x^2} \\ \mathbf{\cos\Big(\arcsin\left(x\right)\Big) =\sqrt{1-x^2} } } \\\\ &=& xy + \sqrt{1-x^2}\sin\Big(\arccos\left(y\right)\Big) \\ && \boxed{ \arccos\left(y\right) = \beta\\ y=\cos(\beta)\\ y^2=\cos^2(\beta)\\ y^2=1-\sin^2(\beta)\\ \sin^2(\beta)=1-y^2 \\ \sin(\beta) =\sqrt{1-y^2} \\ \mathbf{\sin\Big(\arccos\left(y\right)\Big) =\sqrt{1-y^2} } } \\\\ &=& xy + \sqrt{1-x^2}\sqrt{1-y^2} \quad | \quad x=\dfrac{3}{5},\ y=\dfrac{15}{17} \\ &=& \dfrac{3}{5}*\dfrac{15}{17} + \sqrt{1-\dfrac{3^2}{5^2}}\sqrt{1-\dfrac{15^2}{17^2}} \\ &=& \dfrac{45}{85} + \dfrac{4}{5} * \dfrac{8}{17} \\ &=& \dfrac{45}{85} + \dfrac{32}{85} \\ &=& \mathbf{ \dfrac{77}{85} } \\ \hline \end{array}\)

What is the slope of the line that contains the points (-5,4)(-5,7)

Hello Duckielo!

\(m=\frac{y_2-y_1}{x_2-x_1}=\frac{7-4}{-5-(-5)}=\frac{3}{0}=\infty\)

\( The\ slope\ of\ the\ line\ is\color{blue}\ \infty.\)

\(The\ function\ straight\ is\ a\ vertical.\)

  !

For
x > 1, simplify \(\sqrt{ x + \sqrt{2x - 1} } - \sqrt{ x - \sqrt{2x - 1} }\).

\(\text{Let $k= \sqrt{2x - 1},$ or $k^2=2x-1$}\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \sqrt{ x + \sqrt{2x - 1} } - \sqrt{ x - \sqrt{2x - 1} } } \\\\ &=& \sqrt{ x + k } - \sqrt{ x - k } \\\\ &=& \sqrt{ \left( \sqrt{ x + k } - \sqrt{ x - k } \right)^2 } \\\\ &=& \sqrt{ (x+k) -2\sqrt{ (x+k)(x-k) } + (x-k) } \\\\ &=& \sqrt{ 2x -2\sqrt{ x^2-k^2 } } \quad | \quad k^2=2x-1 \\\\ &=& \sqrt{ 2x -2\sqrt{ x^2-(2x-1) } } \\\\ &=& \sqrt{ 2x -2\sqrt{ x^2-2x+1 } } \\\\ &=& \sqrt{ 2x -2\sqrt{ (x-1)^2 } } \\\\ &=& \sqrt{ 2x -2 (x-1) } \\\\ &=& \sqrt{ 2x -2x +2 } \\\\ &=& \mathbf{ \sqrt{ 2 } } \quad | \quad x \geq 1 \\ \hline \end{array}\)

Is it possible cut the obtuse triangle into acute triangular pieces?

The regular price is 59$ , It gets reduced 45%

so we can write:

100%-45%=55% (Current price here we want to find it) 

59$:    x     = y 

x=59*45/100=26.55$

59-26.55=32.45$ Current cost that's what the price it is on sale right now, you would save 26.55 dollars! 

I'm getting an answer of 14/81.

I'm getting that the area of triangle DMT is 30.

im still kind of unsure how you solved the last one (4)? what did you multiply or divide?

1    (3.25 - K) = 5 kg  .890 m/s2        Solve for K

      3.25 - k = 5 (.890)

       3.25-k = 4.45

           k = -1.2 N

2 yes

3 yes

4  mg sin 12  is the force down-plane

    f= mg sin 12 = ma

        g sin 12 = a =2.04 m/s²     (this should make you realize that the acceleration is not dependent on the mass of the object)

All triangles share the altitude.

Let CD be X. Let the shared altitude be Y.

Make some equations!

(1/2) * y(9-x) = 30

(1/2) * y(7-x) = 20

I think from here you pretty much solve for X and Y.

Once you have those values, you can do (1/2) * X * Y for the answer.

Calculator User, not logged in.

If xy = 24, xz = 30, and yz = 15, then find x + y + z.

Hello bigmac!

\(x=\frac{24}{y}\ ;\ \ x=\frac{30}{z}\\ y=\frac{24}{x}\ ;\ \ y=\frac{15}{z}\\ z=\frac{30}{x}\ ;\ \ z=\frac{15}{y}\)

\(x=2y\\ xy=24\\ 2y^2=24\\ \color{blue}y=\sqrt{12}\\ \color{blue}y=2\sqrt{3}\\ \color{blue}x=4\sqrt{3}\)

\(z=\frac{15}{y}=\frac{15}{2\sqrt{3}}=\frac{\sqrt{3}\cdot 5}{2}\\ \color{blue}z=\frac{\sqrt{3}\cdot 5}{2}\)

\(x+y+z=4\sqrt{3}+2\sqrt{3}+\frac{5\sqrt{3}}{2}=\sqrt{3}\cdot \frac{17}{2}\)

  !

your right! I need to be more confident, im alway second guessing and overthink it! thanks!

1.    Remember f= ma ?

           the force net on the object is 3.25 one direction and kinetic frictional force in the OPPOSITE direction

                (3.25 - K) = 5 kg  .890 m/s²        Solve for K

2.   Same thing here   f=ma 

         I THINK what they mean is 'equal and opposite reactions'   and conservation of energy etc

             f = ma    for the 55 kg student

                 = 55 kg (2.5 m/s²⁾     This force will also act BUT in the opposite direction for the 60 kg student

                    55 (2.5) = 60 kg a     Solve for a

3.   conservation of momentum

         Think about a large truck hitting a smaller car....which will have the biggest change in velocity ???

4. You must find the force acting DOWN the plane (NOT the NORMAL force.....we are ignoring friction))

       then    f = ma   

You THINK??  Be a little more confident in your abilities !        Yes, it is "C"