All categories

3 years ago

A=14cm b=9cm c=6cm law of cosines

1 answer:

3 0

A^2= b^2+c^2 - 2bc•cosA

14^2= 9^2 + 6^2 -2x9x6xCosA
196= 81 + 36 - 108xCosA
196- 117= 108CosA
79/108= cos A
0.7315= cosA
A= cos^-1(0.7315)
A= 42.98

SineA/a=sineB/b= sineC/c

Sin42.98/14 = sineB/9
B = sin^-1((9•sin42.98)/14)
B= 25.99

C= 180-(25.99+ 42.98)
C= 111.03

You might be interested in

How do I do number 3

kogti [31]

you simply have to add all of the numbers together, so it'll be 2+5+12+7+1 or 27, therefore the answer is B

5 0

3 years ago

Read 2 more answers

Expand the binomial<br> <img src="https://tex.z-dn.net/?f=%28x-%5Cfrac%7B2%7D%7B5%7D%29%5E%7B2%7D" id="TexFormula1" title="(x-\f

kkurt [141]

Answer:

$(x- \frac{2}{5})^{2} = x^2 -\frac{4}{5}x + \frac{4}{25}$

Step-by-step explanation:

You have two methods to expand this binomial.

Method 1

If you have the expression:

$(x- \frac{2}{5})^{2}$

You can write the expression it in the following way:

$(x-\frac{2}{5})^{2}=(x-\frac{2}{5})(x-\frac{2}{5})$

Then, apply the distributive property:

$(x-\frac{2}{5})(x-\frac{2}{5}) = x^2 -\frac{2}{5}x -\frac{2}{5}x+ (\frac{2}{5})\frac{2}{5}$

Simplify the expression:

$(x-\frac{2}{5})^2= x^2 -\frac{4}{5}x+ (\frac{4}{25})$

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

Method 2

For any expression of the form:

$(a-b)^2$

Its expanded form will be:

$(a-b)^2= a^2 -2ab + b^2$

$a = x$

$b =\frac{2}{5}$

$(x- \frac{2}{5})^{2} = x^2 - 2x\frac{2}{5} + (\frac{2}{5})^2$

$(x- \frac{2}{5})^{2} = x^2 -\frac{4}{5}x + \frac{4}{25}$

3 0

3 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

Saritza will find the area of the rectangle by tiling it with unit squares. The sides: 5/6 and 2/3. The question: Which is the a

AveGali [126]

A rectangle is a 2-dimensional shape with equal opposite sides. The appropriate side length for the unit square to tile Saritza’s rectangle to find the area will be 1/9 yard

<h3>How to find the area of a rectangle</h3>

A rectangle is a 2-dimensional shape with equal opposite sides.

Area of a rectangle = Length * width

Given the following parameters:

length = 5/6 yard

width = 2/3

Find the area

Area = 5/6 * 2/3
Area = 5/9 square units

The appropriate side length for the unit square to tile Saritza’s rectangle to find the area will be 1/9 yard

Learn more on area of rectangle here: brainly.com/question/2607596

3 0

2 years ago

An advertising company designs a campaign to introduce a new product to a metropolitan area of population 3 Million people. Let