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3 years ago

Couldn't you use the law of cosines to find one angle and use the law of sines to find the rest

1 answer:

7 0

Correct. In the case where you are given enough information to use the law of cosines you could in fact then use the law of sines afterwards to find your remaining angle. That being said beware of solutions that don't make a feasible triangle (if you were using the law of cosines you only have one angle, so that means whatever your second angle is that you found using the law of sines can't make your sum go over 180, because you still need some angle left for the last angle).

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Find X in the figure below.

Hitman42 [59]

<h3>Answer:</h3>

x = 3

<h3>Explanation:</h3>

The product of the lengths of segments from the intersection point to the circle is the same for both secants.

... 1×6 = 2×x

... 6/2 = x = 3 . . . . . divide by 2

_____

<em>Comment on secant geometry</em>

Interestingly, this relation is true whether the point of intersection of the secants is inside the circle or outside.

When it is outside, the product is of the distance to the near intersection with the circle and the distance to the far intersection with the circle.

6 0

3 years ago

Pyramid A has a triangular base where each side measures 4 units and a volume of 36 cubic units. Pyramid B has the same height,

omeli [17]

Answer:

The volume of pyramid B is 81 cubic units

Step-by-step explanation:

Given

<u>Pyramid A</u>

$s = 4$ -- base sides

$V = 36$ -- Volume

<u>Pyramid B</u>

$s = 6$ --- base sides

Required

Determine the volume of pyramid B <em>[Missing from the question]</em>

From the question, we understand that both pyramids are equilateral triangular pyramids.

The volume is calculated as:

$V = \frac{1}{3} * B * h$

Where B represents the area of the base equilateral triangle, and it is calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where s represents the side lengths

First, we calculate the height of pyramid A

For Pyramid A, the base area is:

$B = \frac{1}{2} * s^2 * sin(60)$

$B = \frac{1}{2} * 4^2 * \frac{\sqrt 3}{2}$

$B = \frac{1}{2} * 16 * \frac{\sqrt 3}{2}$

$B = 4\sqrt 3$

The height is calculated from:

$V = \frac{1}{3} * B * h$

This gives:

$36 = \frac{1}{3} * 4\sqrt 3 * h$

Make h the subject

$h = \frac{3 * 36}{4\sqrt 3}$

$h = \frac{3 * 9}{\sqrt 3}$

$h = \frac{27}{\sqrt 3}$

To calculate the volume of pyramid B, we make use of:

$V = \frac{1}{3} * B * h$

Since the heights of both pyramids are the same, we can make use of:

$h = \frac{27}{\sqrt 3}$

The base area B, is then calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where

$s = 6$

So:

$B = \frac{1}{2} * 6^2 * sin(60)$

$B = \frac{1}{2} * 36 * \frac{\sqrt 3}{2}$

$B = 9\sqrt 3$

So:

$V = \frac{1}{3} * B * h$

Where

$B = 9\sqrt 3$ and $h = \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9\sqrt 3 * \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9 * 27$

$V = 81$

6 0

3 years ago

Read 2 more answers

Your class is selling boxes of flower seeds as a fundraiser. The total profit p depends on the amount x that your class charges

Arte-miy333 [17]

I cant figure it out sorry :/

3 0

4 years ago

Solve the system of equations using the addition method 2a+3b=-1 ; 3a+5b=-2

Alexeev081 [22]

Answer:

Then, a=-1 and b=-1

Step-by-step explanation:

We need to solve the following system of equations using the adition method:

2a+3b=-1 (i)

3a+5b=-2 (ii)

First, we multiply the equation (i) by -3 and the equation (ii) by 2, so we get:

-6a-9b=3

6a+10b=-4

Adding the two equations we have:

-6a-9b=3

6a+10b=-4

--------------------------

0 + b = -1

Then b=-1. Now, let's substitute the value of b into equation (i):

2a+3b=-1

2a + 3(-1) = -1

2a -3 = -1

2a = 2

a = 1

Then, a=-1 and b=-1

3 0

4 years ago

Read 2 more answers

Help I need to turn this in by 12:30

11Alexandr11 [23.1K]

B- -3

If x is two according to the coordinate points, that means the equation is now -8+3y=-17

Add 8 to each side- 3y=-9

And divide -9 by 3

This gets you y=-3

5 0

3 years ago