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Alexandra [31]

Use photo math. or mathpapa to help you solve the equation

7 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

*PLEASE HELP* *ANY IRRELEVANT ANSWERS WILL BE REPORTED AS SPAM*

drek231 [11]

Answer:

67.307

Step-by-step explanation:

30×25=750

100×80=8000

750+8000=8750

8750/130=67.307... total hits divided by total games

8 0

2 years ago

Read 2 more answers

Ms Kelly puts more than 5 apples in each bag. Write an inequality to represent the number of apples in a bag

Hunter-Best [27]

Y=5+x is the inaquality you are looking for.

3 0

3 years ago

Question 2 A birthday cake in the shape of the right-circular cylinder of radius 15cm and thickness 10cm is cut into smaller pie

finlep [7]

The volume of a piece of the cake will be 588.75 cm³.

The complete question is attached below.

<h3>What is Geometry?</h3>

It deals with the size of geometry, region, and density of the different forms both 2D and 3D.

A birthday cake in the shape of the right-circular cylinder of radius 15 cm and thickness 10 cm is cut into smaller pieces.

One of the piece is shown.

Then the volume of piece of the cake will be

V = (30 / 360) x π x 15² × 10

V = 588.75 cm³

More about the geometry link is given below.

brainly.com/question/7558603

#SPJ1

5 0

2 years ago