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Alenkasestr [34]
1 year ago
6

Consider AABC with vertices A(-2, -3) and B(-6, -6) and C(-8, 5).a) Draw AABC on graph paper. Is this a right triangle? JUSTIFY

your answer.b) Reflect AABC across AC to create AA'B'C'. Find the location of B'. What kind oftriangle is ABB'C:? JUSTIFY your answer.
Mathematics
1 answer:
Fiesta28 [93]1 year ago
6 0

Given data:

The graph of the given points is shown below.

The slope of the line passing through A(-2, -3) and B(-6, -6).

\begin{gathered} m=\frac{-6-(-3)}{-6-(-2)} \\ =\frac{-3}{-4} \\ =\frac{3}{4} \end{gathered}

The slope of the line passing through B(-6, -6) and C(-8, 5).

\begin{gathered} m^{\prime}=\frac{5-(-6)}{-8-(-6)} \\ =-\frac{11}{2} \end{gathered}

The slope of the line passing through A(-2, -3) and C(-8, 5).

\begin{gathered} m^{\doubleprime}=\frac{5-(-3)}{-8-(-2)} \\ =\frac{8}{-6} \\ =-\frac{4}{3} \end{gathered}

The product of the slope of line passing through AB and line passing through AC is,

\begin{gathered} m\times m^{\doubleprime}=\frac{3}{4}\times(-\frac{4}{3}_{}) \\ =-1 \end{gathered}

Thus, the line passing through AB and line passing through AC are perpendicular.

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Step-by-step explanation:

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i have been trying to solve this compound and double angle question please help me find the answer to these question guys​
natali 33 [55]

Answer:

These type of questions are super tricky b/c you have to remember all the different versions of the identities, and then they put the question in some odd form,  I feel like this should land math professors in jail , for dishonesty , b/c it's really a form of "how tricky can I make a question and still have a way to solve it"   anyway,

Step-by-step explanation:

a)

next the question asks   1-cos 2A   and this is total abuse of notation.   the way this should be written is   1- cos( 2A)  so we know that the A is part of the cosine functions input... btw.. in any computer program,  it would never ever let you get away with that top form of the expression.  :/   anyway... I keep ranting.. huh... sorry  :P

1-cos(2A) is an odd form of the identity  1/2(1-cos(2A) = sin^{2}(A)  the 1/2 is missing but we can add that pretty easy, we just have to remember to take it out too. I usually forget to do that. and my professor marks me off completely,  totally wrong, but I just miss one small thing  :/  anyway....

our 1-cos(2A) needs the 1/2 added to it.  or if we move that 1/2 to the other side it looks like  2*sin^{2}(A)  = 1-cos(2A)  and this is that "odd" from of the identity that I was talking about.  

next let's deal with sin(2A)  it has an identity of  2 sin(A)cos(A) which is really nice for us b/c it will cancel out the 2 in then numerator for us, nice !

now our fraction looks like  [2* sin^{2}(A)] / 2 sin(A)cos(A)

so cancel out one of the sines

2*sin(A) / 2 cos(A)

cancel the 2s

Sin(A) / Cos(A) = Tan(A)

nice  it worked out  :P

b)

by the above that we just worked out, then

Tan(15) = Sin(15) / Cos(15)

I had to look up what sin of 15 is b/c it's not one of those special angles but it does have an exact form of

Sin(15) = (√3 - 1) / 2√2

Cos(A) = (√3 + 1) / 2√2

you can use rule of Cos(A-B) = Cos(A)Cos(B)+Sin(A)Sin(B) to get the above and a similar rule for Sin(A-B)

back to our problem,  the 2√2 will cancel out

then we have

Tan(15) =  (√3 - 1) /(√3 + 1)

in the form that is above that's exact, the roots could be approximated but i'll just leave that in the form that is exact.  Most math professors like that form.  

 

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