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Goshia [24]
2 years ago
5

Triangle △ABC is reflected across line n to create △ A'B'C', What is the measure of ∠C?

Mathematics
1 answer:
Ilya [14]2 years ago
6 0

The measure of ∠C is 54°.

<h3>To solve the problem :</h3>

let us assume that line n is a mirror.

∴ ∠A = ∠A' = 59°

∠B = ∠B' = 67°

∠C = ∠C'

According to property of triangles :

Sum of all angles of a triangle is 180°.

∴ A + B + C = 180°

59 + 67 + C = 180

C = 180 - (59 +67)

C = 54°

<h3>∴ The measure of ∠C is 54°.</h3>

To know more about triangles refer to :

brainly.com/question/1620555

#SPJ2

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It's not difficult to compute the values of A and B directly:

A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}
A=\tan^{-1}(\sin\theta)-\dfrac\pi4

B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt
B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}
B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2

Let's assume 0, so that |\csc\theta|=\csc\theta.

Now,

\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}
\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}
\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)
\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)

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What is the argument of -5\sqrt{3}+ 5i?
Shtirlitz [24]

Answer:

150 degrees

Step-by-step explanation:

Graphing the complex number we see the angle terminates in the second quadrant. This means the argument, the angle, will be between 90 degrees and 180 degrees.

So if we create a right triangle with that point after graphing it. We see the height of that triangle is 5 because that is the imaginary part. The base of that triangle has length 5\sqrt{3}. The problem is this doesn't give us any part of the angle we want, but it does give us the complementary of the part of the angle that is in second quadrant.

Let's find the complementary angle.

So the opposite side of the complementary angle is 5.

The adjacent side of the complementary angle is 5\sqrt{3}.

\tan(\theta)=\frac{5}{5\sqrt{3}}

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Step-by-step explanation:

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