hectictar

What is the degree-measure of the acute angle formed by extending sides AB and ED of regular nine-sided polygon ABCDEFGHI until these extensions meet?

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Let the point of intersection of lines AB and ED  be  P .

sum of interior angle measures of nonagon ABCDEFGHI  =  180°(9 - 2)  =  1260°

ABCDEFGHI is a regular polygon, so each interior angle has the same measure.

measure of each interior angle of ABCDEFGHI  =  1260° / 9   =   140°

sum of interior angle measures of heptagon APEFGHI  =  180°(7- 2)  =  900°

 m∠P + 840°   =   900°

 m∠P   =   60°

If  \(\sin^2(\frac{\pi}{9})+\sin^2(\frac{2\pi}{9})+\sin^2(\frac{3\pi}{9})+\sin^2(\frac{4\pi}{9})=\frac94\) , evaluate

\(\cos^2(\frac{\pi}{9})+\cos^2(\frac{2\pi}{9})+\cos^2(\frac{3\pi}{9})+\cos^2(\frac{4\pi}{9})\\ \text{____________________________________________} \)  .

By the Pythagorean identity,

\(\sin^2\theta+\cos^2\theta=1\)       so       \(\cos^2\theta=1-\sin^2\theta\)

So

\(\ \qquad\cos^2(\frac{\pi}{9})+\cos^2(\frac{2\pi}{9})+\cos^2(\frac{3\pi}{9})+\cos^2(\frac{4\pi}{9})\\ =\\ \qquad (\,1-\sin^2(\frac{\pi}{9})\,)+(\,1-\sin^2(\frac{2\pi}{9})\,)+(\,1-\sin^2(\frac{3\pi}{9})\,)+(\,1-\sin^2(\frac{4\pi}{9})\,)\\ =\\ \qquad \phantom{(\,}1-\sin^2(\frac{\pi}{9})\phantom{\,)}+\phantom{(\,}1-\sin^2(\frac{2\pi}{9})\phantom{\,)}+ \phantom{(\,}1-\sin^2(\frac{3\pi}{9})\phantom{\,)}+\phantom{(\,}1-\sin^2(\frac{4\pi}{9})\phantom{\,)}\\ =\\ \qquad4-\sin^2(\frac{\pi}{9})-\sin^2(\frac{2\pi}{9})-\sin^2(\frac{3\pi}{9})-\sin^2(\frac{4\pi}{9})\\ =\\ \qquad4-(\sin^2(\frac{\pi}{9})+\sin^2(\frac{2\pi}{9})+\sin^2(\frac{3\pi}{9})+\sin^2(\frac{4\pi}{9}))\\ =\\ \qquad4-\frac94\\ =\\ \qquad\frac{16}{4}-\frac94\\ =\\ \qquad\frac74 \)

We can check this answer with the calculator by entering:

(cos(pi/9))^2+(cos(2pi/9))^2+(cos(3pi/9))^2+(cos(4pi/9))^2

which does equal  1.75  

26.

Since  AB  is tangent to the circle at  A ,  m∠PAB  =  90°

Since  PA  and  PC  are both radii of circle P, they are the same length.

So    the length of PC  =  x    and    the length of PB  =  8 + x

And by the Pythagorean Theorem...

x² + 12²  =  (8 + x)²

x² + 144  =  (8 + x)(8 + x)

x² + 144  =  64 + 16x + x²

144  =  64 + 16x

80  =  16x

5  =  x

(I think that  PB  is not a secant.)

46.

csc is the reciprocal of sin.

If he named 1 number, then the last number he named would be 15

If he named 2 numbers, the last number he named would be 15 + 6

If he named 3 numbers, the last number he named would be 15 + 6 + 6

If he named 4 numbers, the last number he named would be 15 + 6 + 6 + 6  which is  15 + 6(4 - 1)

So... if he named  n  numbers, the last number he named would be  15 + 6(n - 1)

They tell us the last number he named was  363 . So..

363  =  15 + 6(n - 1)    Subtract  15  from both sides.

348  =  6(n - 1)            Divide both sides by  6 .

58  =  n - 1                  Add  1  to both sides.

59  =  n

40)

You are given two triangles and the information that the three pairs of corresponding angles are congruent. What other information would guarantee that the triangles are congruent?

D: One pair of corresponding sides has the same measure.

43)

If minor arc AC = 96°, what is the measure of ∠ABC?

Here's how I have to think of these....

Draw the angle that forms minor arc AC...its measure is 96° .

Even though it wasn't specifically stated in the question, I assume that lines AB and BC are tangent to the circle. That means line AB and line BC each forms a right angle with the line drawn from the center of the circle to point A and point C respectively.

The sum of the angle measures in every quadrilateral is  360° . So....

m∠ABC  =  360° - 90° - 90° - 96°  =  180° - 96°  =  84°

32)

Since the sum of the angle measures in every triangle is  180°,

and we know that two of the angles are  90°  and  45°.....

the measure of the third angle   =   m∠A   =   180° - 90° - 45°   =   45°

Since  m∠A  =  m∠C ,  triangle  ABC  is isosceles and   AB  =  BC

By the Pythagorean Theorem...

(AB)² + (BC)²  =  n²

                                      We can replace  AB  with  BC  since they are the same length.

(BC)² + (BC)²  =  n²

                                      Combine like terms.

2(BC)²  =  n²

                                      Divide both sides of the equation by  2 .

(BC)²  =  n² / 2

                                      Take the positive square root of both sides.

BC  =  \(\sqrt{\frac{n^2}{2}}\)

                                      Simplify.

BC  =  \(\frac{\sqrt{n^2}}{\sqrt2}\)

BC  =  \(\frac{n}{\sqrt2}\)

                                      Rationalize denominator.

BC  =  \(\frac{n\sqrt2}{2}\)

To solve the equation for x....

\(|x|=x^2+x-3\\~\\ x=\pm(x^2+x-3)\\~\\ \begin{array}\ x=+(x^2+x-3)&\qquad\text{or}\qquad& x=-(x^2+x-3)\\~\\ x=x^2+x-3&&x=-x^2-x+3\\~\\ 0=x^2-3&&0=-x^2-2x+3\\~\\ 3=x^2&&x^2+2x-3=0\\~\\ \pm\sqrt3=x&&(x+3)(x-1)=0\\~\\ x=\sqrt3\quad\text{or}\quad x=-\sqrt3&\qquad\text{or}& x=-3\quad\text{or}\quad x=1 \end{array} \)

Now we need to test each possible solution to see if it makes the given equation true:

\(|\sqrt3|\,\stackrel?=\,\sqrt3^2+\sqrt3-3\\~\\ \sqrt3\,\stackrel?=\,3+\sqrt3-3\\~\\ \sqrt3\,\stackrel?=\,\sqrt3\)

Yes, so  \(x=\sqrt3\)  is a solution.

\(|-\sqrt3|\,\stackrel?=\,(-\sqrt3)^2-\sqrt3-3\\~\\ \sqrt3\,\stackrel?=\,3-\sqrt3-3\\~\\ \sqrt3\,\stackrel?=\,-\sqrt3 \)

No, so  \(x=-\sqrt3\)  is not a solution.

\(|-3|\,\stackrel?=\,(-3)^2-3-3\\~\\ 3\,\stackrel?=\,9-3-3\\~\\ 3\,\stackrel?=\,3\)

Yes, so  \(x=-3\)  is a solution.

\(|1|\,\stackrel?=\,1^2+1-3\\~\\ 1\,\stackrel?=\,1+1-3\\~\\ 1\,\stackrel?=\,-1\)

No, so  \(x=1\)  is not a solution.

So the solutions are  \(x=\sqrt3\)  and  \(x=-3\)  .

The vector in the opposite direction of  \(\vec u\)  is  \(-\vec u\)  .

The unit vector in the same direction of  \(\vec u\)  is  \(\frac{\vec u}{|\vec u|}\)  which is  \(\frac{\vec u}{7}\)  .

So the unit vector in the opposite direction of  \(\vec u\)  is  \(-\frac{\vec u}{7}\)  .

(a + b)(a + b)  =  (a)(a) + (a)(b) + (b)(a) + (b)(b)  =  a² + ab + ab + b²  =  a² + 2ab + b²

So....   a² + 2ab + b²  =  (a + b)(a + b)

And if

(a² + 2ab + b²)(a + b)  =  125      then....

(a + b)(a + b)(a + b)  =  125

(a + b)³  =  125

a + b  =  \(\sqrt[3]{125}\)

a + b  =  5