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2 years ago

Convert angle radians to degrees

Mathematics

1 answer:

boyakko [2]2 years ago

5 0

Answer:

Step-by-step explanation:

$\frac{\theta }{rad}=\frac{360}{2\pi }\\ \\ \theta =\frac{180rad}{\pi }$

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How much of a circle does a 100-degree angle turn through?

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2 years ago

What is -2(5+6m) + 16 = -90 ?<br> Can you please show with detail

GarryVolchara [31]

Answer:

m = 8

Step-by-step explanation:

-2(5 + 6m) + 16 = -90

Start by distributing -2 inside the parentheses. This means you will multiply everything inside the parentheses (5 + 6m) by -2.

-10 - 12m + 16 = -90

Simplify the left side of the equation by combining like terms.

6 - 12m = -90

Subtract 6 from both sides of the equation to isolate the term containing the variable m and to move all other terms to the other side (right) of the equation.

-12m = -96

Divide both sides of the equation by -12 to isolate and solve for m.

m = 8

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2 years ago

Write a paragraph explaining how to convert written words into algebraic expressions.

amid [387]

Step-by-step explanation:

They want you to write a paragraph about addition, subtraction, multiplication, and division.

5 0

3 years ago

Need some help please

katovenus [111]

Answer:

the answer is 40.1 cm

Step-by-step explanation:

$\cos( \alpha ) = adjacent \div hypotenuse$

The adjacent in the triangle is 23cm and the Alpha is 55 degrees, so just insert the formula to find the adjacent which is x .

8 0

2 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago