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Mathematics 16 Online
OpenStudy (anonymous):

find ae. round the answer to the nearest tenth. a. 9.7 b. 13.5 c. 16.1 d. 17.3

OpenStudy (anonymous):

OpenStudy (anonymous):

@pitamar

OpenStudy (anonymous):

Do you want guidance or the answer?

OpenStudy (anonymous):

whichever

OpenStudy (anonymous):

as long as its fast im fine with whichever

OpenStudy (anonymous):

cuz im being timed

OpenStudy (anonymous):

Do you know the law of cosines?

OpenStudy (anonymous):

somewhat not reallly

OpenStudy (anonymous):

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

OpenStudy (anonymous):

it is an extension of the pythagorean theorem to all triangles, not just right angle triangles

OpenStudy (anonymous):

somewhat not reallly

OpenStudy (anonymous):

to use it, we have to figure out AB and angle ABE. we are told BE so that's all we need

OpenStudy (anonymous):

to find AB we can use sin() function. can you find AB?

OpenStudy (anonymous):

do i need a calculator?

OpenStudy (anonymous):

31 degrres

OpenStudy (anonymous):

we know that \(\sin( \angle ABC) = \frac{AC}{AB}\)

OpenStudy (anonymous):

which becomes $$\sin(31^\circ) = \frac{6}{AB}$$

OpenStudy (anonymous):

can you solve this one for AB?

OpenStudy (anonymous):

3.06?

OpenStudy (anonymous):

arm.. let's just rearrange this. we can multiply both sides by AB: $$ \sin(31^\circ) = \frac{6}{AB} \implies AB \cdot \sin(31^\circ) = 6 $$ and then divide by sin(31): $$ AB = \frac{6}{\sin(31^\circ)} $$calculating it I get 11.65

OpenStudy (anonymous):

now the last thing we need is angle ABE. can you find it?

OpenStudy (anonymous):

dont know how

OpenStudy (anonymous):

we know that $$\angle CBA + \angle ABE + \angle EBD = 180^\circ $$ right? they all together make a flat angle. We are told CBA and EBD. what are they?

OpenStudy (anonymous):

31 and 13

OpenStudy (anonymous):

Right, so we plug that in: $$\angle CBA + \angle ABE + \angle EBD = 180^\circ \\ 31^\circ + \angle ABE + 13^\circ = 180^\circ $$ Can you move the numbers to the right and find \( \angle ABE\)?

OpenStudy (anonymous):

136? idk

OpenStudy (anonymous):

yes, exactly

OpenStudy (anonymous):

So, we know $$ AB=11.65\\ BE = 7 \\ \angle ABE = 136^\circ $$Now let's plug it in the formula. we'll say: $$ a = AB = 11.65 \\ b = BE = 7 \\ \alpha = \angle ABE = 136^\circ $$ and our formula is: $$ c^2 = a^2 + b^2 - 2ab \cdot \cos(\alpha) \\ c^2 = (11.65)^2 + 7^2 - 2 \cdot(11.65) \cdot (7) \cdot \cos(136^\circ) $$can you calculate this?

OpenStudy (anonymous):

no ill get lost in them lol

OpenStudy (anonymous):

cmon.. try

OpenStudy (anonymous):

you can type it in here if you prefer: http://www.wolframalpha.com/

OpenStudy (anonymous):

idk how like ima forget all the numbers calculationg it

OpenStudy (anonymous):

type (11.65)^2 + 7^2 - 2*11.65*7*cos(136 degrees)

OpenStudy (anonymous):

you see how it is actually the same expression? just textually. what do you get?

OpenStudy (anonymous):

i got 302

OpenStudy (anonymous):

correct $$ c^2 = 302$$ we want \(c\) not \(c^2\) so we take root: $$ c = \sqrt{302} = ? $$

OpenStudy (anonymous):

17.37

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

thx!!

OpenStudy (anonymous):

np Just remember that in laws of cosines the angle has to be between the two sides you use, like we had in our case

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