sine and cosine law

[true forge]

can someone pls help me w 5e and 5f. ive been doing the previous questions and im still very confused

[true forge]

pls someone help coz i have my math culminating tmrw and i think i alr failed day 1 and 2

[opal cedar]

\section*{5e) $\begin{aligned} &2x + y = 5 \ &x - 3y = 13 \end{aligned}$}

[pastel girder]

@opal cedar

[opal cedar]

@true forge
First, which variable, and in what equation, do you want to isolate?

[true forge]

can i isolate the y in the first equation

[opal cedar]

And what do you get?

[true forge]

is it y=5-2x?

[glass musk]

yes

and then you fill in that y in the other equation

[true forge]

so is it x - 3 (5-2x) = 13?

[glass musk]

yep

then you solve for x

and then you fill that answer back into the other equation

[true forge]

ok. so x = 4? then i sub it into the equation

[glass musk]

1 sec let me calculate that lol

yes x= 4 you're correct

[true forge]

what equation would i sub x into?

the first one?

[glass musk]

2x + y =5 to find y

[true forge]

ok so y =3? and whenever i find the value of x for a problem like this im doing, would i always sub it into the first equation to find y?

[glass musk]

y = -3

[true forge]

wait why -3

[glass musk]

y = 5 -2x <=> y = 5 - 2(4) <=> y = -3

you bring the 2x to the other side so it "becomes" -2x

[true forge]

oh i thought i had to sub x into 2x+y=4

[glass musk]

yes that's correct but you may have forgotten the '-' symbol when "replaced" the 2x

so 2x+y= 5 (we bring the 2x to the other side) y=5-2x (now we fill in the value for x which is 4) and we get y= 5-2(4)

[true forge]

okay thank u

[glass musk]

np

if you need any help with f feel free to ping me

[true forge]

@glass musk can u help pls. i think i messed up smth w f. for y i got -4 but i check the answer sheet and apparently its 4/3

[glass musk]

ok one sec let me check

[true forge]

this is what i did

[glass musk]

ok so x-y=-4 <=> x=y-4 (now we have a value for x and we plug that value into x+2y=0) x+2y =0 <=> (y-4) + 2y =0 <=> y-4+2y= 0 <=> 3y-4 =0 <=> 3y=4 <=> y= 4/3

[true forge]

so it seems like i ended up rewriting the wrong equation. but how do ik which is the right equation to rewrite?

[glass musk]

I always pick the easiest one

so for example if I had 2x+3y=0 and x+3y=5 I'd begin with the 2nd one so x= 5-3y

it doesn't actually matter which one you pick as long as you're not making mistakes you'll get the right answer

just a matter of how long it will take

e.g. it's easier to calculate with x=4 rather than y = 7/3

btw the answer for f:

||now that we have a value for y we can plug it back into the first equation to solve x . So x=y-4 <=> x= 4/3 -4 <=> x= 4/3- 12/3 <=> x = -8/3 and y=4/3||

[true forge]

ohh ok

[true forge]

thank you

[glass musk]

quick tip before you start calculation and jumping right into the question take a few seconds to analyse it, think about 'what would be the easiest way to answerthis question'

[true forge]

okay i will thanks

[glass musk]

trust me I have made the mistake of not thinking and just doing a lot on math test and it seriously screwed me over multiple times, it will take longer to answer the question wrong and doing it all over again,rather than taking your time and getting it right

[true forge]

thanks for the advice

[glass musk]

np

[keen wyvern]

bro it is so easy

[opal cedar]

It might be easy for you, but not for others. If you have nothing helpful to say, please don't be condescending to people who are trying to learn.

[true forge]

can someone help me w question d pls. ik its alr solved but i wasnt the one who did it. my friend gives me the test before i do it

i dont get how im supposed to know whcih sides are a b and c

sine and cosine law

[opal cedar]

,w 18^2 + 13^2

,w 21^2

[pastel girder]

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Requested by en.ratham

[opal cedar]

@true forge
Which part of the solution do you not get?

[true forge]

kinda the whole thing

idk whcih places the numbers are supposed to go into the formula

[opal cedar]

$$c^2 = a^2 + b^2 - 2ab \cos(\mathrm C)$$
$$\begin{aligned}
&c = 21 \
&a = 18 \
&b = 13 \
&\mathrm{C} = \theta \end{aligned}$$

In general, angle C is opposite to side c. a and b represent the other two sides.

[pastel girder]

@opal cedar

[opal cedar]

$$c^2 = a^2 + b^2 - 2ab \cos(\mathrm C)$$
$$\begin{aligned}
&c = 21 \
&a = 18 \
&b = 13 \
&\mathrm{C} = \theta \end{aligned}$$
Plug everything into the formula:
$$21^2 = 18^2 + 13^2 - 2(18)(13)\cos(\theta)$$

[pastel girder]

@opal cedar

[true forge]

but how do ik what the angles and side are if it doesnt say it on the pic

is c always the bottom side?

[opal cedar]

First, your only angle of interest is C, or theta.

[opal cedar]

Then, you can say that c is the side opposite to the angle of interest.

[opal cedar]

Finally, a and b are the two other sides (picked however you want). You also could have solved this with a = 18 and b = 13.

[true forge]

ohhhh i see

ima try to solve it now

wait i have another question. why do i have to switch the -468costheta with the 441

[opal cedar]

People don’t like dealing with negative numbers. Also, isolating cos can make it easier to solve for the angle.

[true forge]

ohh ok so its optional then?

[opal cedar]

You can use any method to solve for theta once you have written the Cosine Rule.

[true forge]

okay

i got 84 idk how they got 83

thanks for the help @opal cedar