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Given a triangle with C=108°, a=10, b=6.5, find side c

$$c^2=a^2+b^2-2ab*\cos(108\ensurement{^{\circ}}) \\
c^2 = 10^2+6.5^2-2*10*6.5*\cos(108\ensurement{^{\circ}}) \\
c^2 = 100 +42.25 - 130 * (-0.30901699437) \\
c^2 = 142.25 + 40.1722092687\\
c^2 = 182.422209269 \quad | \qaud \sqrt{} \\
c= 13.5$$

The person consumes

5.4x10^5 kcal / 3.7x10^6 kcal = 1.459 x 10^-1   =  about .1459 barrels per day

So

365 x .1459 =  about 53.2535 barrels per year

Using the Law of Cosines, we have

C^2 =

 $${{\mathtt{10}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{6.5}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{6.5}}\right){\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{108}}^\circ\right)} = {\mathtt{182.422\: \!209\: \!268\: \!75}}$$

C =

 $${\sqrt{{\mathtt{182.422\: \!209\: \!268\: \!75}}}} = {\mathtt{13.506\: \!376\: \!615\: \!093\: \!701\: \!7}}$$

Well if you are given 1 side and the opposite angle and one other thing then you can use sine rule to find the 4th thing.

If you are given 2 sides and the angle between them OR if you are given 3 sides and you need to find something else then you would use Cosine Rule.

I think that those instructions will work each time  

Sometimes you need to do more than one step and sometimes you could use either.

Chris would you like to comment on this one?

Hi I am not really sure of this question either so can somebody please help. My math teacher gave it to me and it's due next week! o_0

Here's the Venn diagram for the situation....

The percent surveyed that had a cell phone was 234 / 528 = about 44.3%

The percent that owmed only a cell phone was  55 / 528 = about 10.4%

Thanks Chris :)

Yep you are right - it couldbe imperial minims that are needed.  LOL  :))

You know, as a member and the poster of a question you can give points.   :)

Mmm

I have always been a bit 'wishy' about these terms too.

Yeah...what she said.....

Very nice, Melody...!!

Draw a right angled triangle

put x as an acute angle and mark th opposite side 1 and the hypotenuse as 3

find the other side using pythagoras theorum.

read off cosx

s=80   a=9.8     u=0    t= ??

None of those fromulas have got just  s  a u and t   so maybe we will use 2 formulas

No 4

$$\\v^2=u^2+2as\\
v^2=0+2*9.8*80\\
v^2=1568\\
v=39.6m/s$$

Now I will use no1

$$\\v=u+at\\
39.6=0+9.8*t\\
t=39.6/9.8\\
t=4.04 seconds$$

---------------------------------------

 You can also use calculus to solve this.

$$\\ \ddot{y} = -9.8\\
\dot{y}=-9.8t\\
y=-9.8/2*t^2+80\\
y=-4.9t^2+80\\
when y=0
0=-4.9t^2+80\\
4.9t^2=80\\
t^2=80/4.9\\
t=4.04 seconds$$

Great answer saseflower :)

$${\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{10}}}}\right)} = {\mathtt{0.960\: \!264\: \!500\: \!792\: \!218}}$$

3/2 ÷ 1/6

Use the recipricle and multiply

3/2 x 6/1

Multiply across

18/2

Simplify

9

Hope this helps :)

From the first observation, we have

tan 36 = h / x      where h is the height of the tree and x is the distance the tree is from the first observation point

And solving for x, we have  x = h / tan 36

And from the second observation, we have

tan 65 = h / (x - 20)    and solving for x, we have

(x - 20) tan 65 = h     

x - 20 = h /tan 65 

x = h / tan 65 + 20 

And setting the "x's" equal, we have

h / tan 36  = h/tan 65 + 20    rearrange

h / tan 36 - h / tan 65  = 20    multiply both sides by the common denominator, (tan 65 * tan 36)

h ( tan 65 - tan 36) = 20 (tan 65 * tan 36)

h = 20(tan (65) * tan (36) ) / (tan( 65) - tan (36)) = about 21.98 ft

Here's a pic.....

You have given no initial units

you can also use this symbol       ^

It is because the forum is getting increasingly well know and almost every child in the world studies maths and has questions.  

The answerers here are not paid so it is amazing how many highly qualified people donate their time to answerer questions but we simply cannot get through them all.

It is great that so many less qualified people also get in and answer questions within their capabilities.

This is what I always wanted for this forum.   

It is a sad fact that the forum is flooded with questions and we simply cannnot answer them all!

Members who become known to us and have good attitudes will always get priority.   

Thank you, CPhill! Proportions are my favorite thing to solve in math :)

Nice, saseflower....!!!