Can you help me plz

The terminal side of an angle, Theta, whose sine value is One-half

Kazeer [188]

Theta has a reference angle of 30° and is in Quadrant I or II.

Sin(theta) = ½

Basic angle: 30

<h3>What is the reference angle?</h3>

The acute angle between the terminal arm/terminal side and the x-axis. The reference angle is always positive. In other words, the reference angle is an angle sandwiched between the terminal side and the x-axis.

Angles: 30,

180-30 = 150

Because sin is positive in quadrants 1 and 2.

To learn more about the terminal side visit:

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4 0

2 years ago

Given points A (0,0), B (4,0), C (0,2), D (0,0), E (2,0), and F (0,1), which of the following proves that

damaskus [11]

Similar triangles may or may not have equal side lengths

Triangles ABC and DEF are similar by SSS theorem

<h3>How to determine the similarity statement</h3>

The coordinates of the two triangles are:

A (0,0), B (4,0), C (0,2), D (0,0), E (2,0), and F (0,1)

Calculate the lengths of both triangles using the following distance formula

$d = \sqrt{(x_2 -x_1)^2 + (y_2 -y_1)^2}$

For triangle ABC, we have:

$AB = \sqrt{(0 -4)^2 + (0 -0)^2} = 4$

$BC = \sqrt{(4 -0)^2 + (0 -2)^2} = 2\sqrt 5$

$CA = \sqrt{(0 -0)^2 + (0 -2)^2} = 2$

For triangle DEF, we have:

$DE = \sqrt{(0 -2)^2 + (0 -0)^2} = 2$

$EF = \sqrt{(2 -0)^2 + (0 -1)^2} = \sqrt 5$

$FD = \sqrt{(0 -0)^2 + (1 -0)^2} = 1$

Divide corresponding sides of the triangles to calculate the scale factor k

$k = \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}$

This gives

$k = \frac{4}{2} = \frac{2\sqrt 5}{\sqrt 5} = \frac{2}{1}$

Evaluate the quotients

$k = 2 = 2 = 2$

Since the quotients are equal, then the triangles are similar

Hence, triangles ABC and DEF are similar by SSS theorem

Read more about similar triangles at:

brainly.com/question/14285697

3 0

2 years ago

The rectangle below has an area of 1 8 x 3 18x 3 square meters and a length of 2 x 2 2x 2 meters.

motikmotik

Answer:

$\text{The width of rectangle}=9x\text{ meters}$

Step-by-step explanation:

We have been given that the rectangle has an area of $18x^3$ square meters and a length of $2x^{2}$ . We are asked to find the width of rectangle.

Since we know that area of a rectangle is width times length of the rectangle, so we can find width of our given rectangle by dividing given area by length of rectangle.

$\text{Area of rectangle}=\text{Width of rectangle *Length of the rectangle}$