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Afina-wow [57]
3 years ago
14

Help me with the diagram please!!!

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
5 0

Answer:

(B) 30

Step-by-step explanation:

Imagine you drew a line from Point T until it touched Line PR. Let's call that point where it touched Line PR "Point Z".

That line (called Line TZ) would be perpendicular to PR, forming a 90 degree angle.

Now, TZW is a triangle.

To find x, we need to find the angle measurment of Angle ZTW.

This is where we use the hexagon.

A hexagon's interior angle sum is 720, meaning each interior angle is equal to 120 degrees. So Angle UTS would equal 120 degrees.

However, Line TZ bisects that 120 degree angle, so Angle ZTW would equal 60 degrees (because 120/2 = 60).

Now we have two angles of the triangle: 90 & 60.

A triangle's interior angle sum is 180.

Add 90 & 60, which is 150, and subtract 150 from 180.

The result is 30, which is the angle measurement of x.

Hope it helps (●'◡'●)

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Area of a triangle is

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Find the solutions of the following trigonometric equation in the interval [0,2π). tan^2x+secx=1
Citrus2011 [14]
\tan ^2x+\sec x=1

Given the next trigonometric identity:

\begin{gathered} \tan ^2x+1=\sec ^2x \\ \text{ Or} \\ \tan ^2x=\sec ^2x-1 \end{gathered}

Substituting this identity into the equation:

\sec ^2x-1+\sec x=1

Subtracting 1 at both sides of the equation:

\begin{gathered} \sec ^2x-1+\sec x-1=1-1 \\ \sec ^2x+\sec x-2=0 \end{gathered}

Replacing with:

\begin{gathered} y=\sec x \\ y^2=\sec ^2x \end{gathered}

we get:

y^2+y-2=0

Applying the quadratic formula with a = 1, b = 1 and c = -2:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}

Recalling that y = sec(x), then we have two options:

\begin{gathered} \sec x=1 \\ \text{and} \\ \sec x=-2 \end{gathered}

By definition:

\sec x=\frac{1}{\cos x}

Therefore, the first option is:

\begin{gathered} \frac{1}{\cos x}=1 \\ (\frac{1}{\cos x})^{-1}=1^{-1} \\ \cos x=1 \end{gathered}

In the interval of x [0,2π), the solution to this equation is 0.

Now, considering the second option:

\begin{gathered} \frac{1}{\cos x}=-2 \\ (\frac{1}{\cos x})^{-1}=(-2)^{-1} \\ \cos x=-\frac{1}{2} \end{gathered}

In the interval of x [0,2π), the solutions to this equation are 2π/3 and 4π/3.

In summary, the solutions to tan^2⁡(x) + sec(x) = 1 are:

x=0,\frac{2\pi}{3},\frac{4\pi}{3}

5 0
1 year ago
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