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2 years ago

Point A is located at (5, 10) and point B is located at (20, 25).

1 answer:

8 0

Let's say the point is C, so C partitions AB into two pieces, where AC is at a ratio of 3 and CB is at a ratio of 7, thus 3:7,

$\bf \left. \qquad \right.\textit{internal division of a line segment}
\\\\\\
A(5,10)\qquad B(20,25)\qquad
\qquad 3:7
\\\\\\
\cfrac{AC}{CB} = \cfrac{3}{7}\implies \cfrac{A}{B} = \cfrac{3}{7}\implies 7A=3B\implies 7(5,10)=3(20,25)\\\\
-------------------------------\\\\
{ C=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}\\\
-------------------------------$

$\bf C=\left(\cfrac{(7\cdot 5)+(3\cdot 20)}{3+7}\quad ,\quad \cfrac{(7\cdot 10)+(3\cdot 25)}{3+7}\right)
\\\\\\
C=\left( \cfrac{35+60}{10}~~,~~\cfrac{70+75}{10} \right)\implies C=\left(\cfrac{95}{10}~~,~~\cfrac{145}{10} \right)
\\\\\\
C=\left( \cfrac{19}{2}~~,~~\cfrac{29}{2} \right)\implies C=\left( 9\frac{1}{2}~~,~~14\frac{1}{2} \right)$

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x = -20
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do it again for the other angle measurement
60+60+40=180

therefore, angle A is 60°

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2 years ago

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2 years ago

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2 years ago

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

2 years ago

Due tomorrow please help :)

coldgirl [10]

Okay, so your first one, number six. You have two shapes there, a cone and a circle. Now you have a full cone and a half of a circle. Let's start with the cone. The formula for the volume of a cone is πr²h/3. You have your height, that is 12, so get your paper out and plug that in. H = 12. Now, your radius is where it gets a bit tricky. Have you ever heard of the Pythagorean theorem? It's a²+b²=c². Look at the cone, but look at it like it's a two-dimensional shape. You see the triangle in the middle, ignore the rest and pretend it's only that triangle. You now have a right triangle, You know your height is 12, and by looking at the triangle adjacent to it, and know it's equal, you know that your hypotenuse (the longest side of your triangle) is 15. So plug that into the Pythagorean theorem. Now you have 12²+b²=15². Use algebra to work that out, and you're left with 9. So your radius is 9. Now plug that into your volume of a cone equation. The volume of the cone is about 1017.88. Now find the circle or sphere. The formula is 4/3πr³. Just plug in your radius and get your answer, which is 3053.63. Now cut that in half, as you've only got half a circle. Now you've got 1526.82. Now add that to the volume of your cone, and you've got your final answer. 2544.7, or about that. I used π on a calc, and not 3.14, so it might be a little different.

Do you think you can do the next one? Just look at your triangle in there. You've already got one side and a hypotenuse, so plug it into the Pythagorean theorem and get your last side. Then look at the little marks on there and you can see if places are the same. If you need help with it don't hesitate to ask. I'm staying up all night anyway. Homeschooled and behind, so I'm not going anywhere.

7 0

2 years ago