Help please asap thanks

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

Find the measure of angle x

Triss [41]

Answer:

it is heptagon and sum of interior angles of heptagon is 900 °

123°+131°+ 125°+124°+x 129°+132° = 900°

764° +x = 900 °

x = 900°-764°

x = 136°

4 0

3 years ago

Can you please help me

max2010maxim [7]

Here’s the answer and each step I took to solve it. With the example I made on the side just continue to repeat that for each one to get the answers. You just replace that one number for the X to take the place for the next X value as you go down the table solving them.

7 0

3 years ago

Find the linear equation of the plane through the point (2,4,9) and parallel to the plane x+4 y+5 z+4 =0.

storchak [24]

Y = x + 5A linear equation (in slope-intercept form) for a line perpendicular to y = -x + 12 with a y-intercept of 5.y = 1/2x - 5Convert the equation 4x - 8y = 40 into slope-intercept form.y = -1/2x + 5A linear equation (in slope-intercept form) which is parallel to x + 2y = 12 and has a y-intercept of 5.3x - y = -5A linear equation (in standard form) which is parallel to the line containing (3, 5) and (7, 17) and has a y-intercept of 5.y = -3x + 1A linear equation (in slope-intercept form) which contains the points (10, 29) and (-2, -7).y = -5A linear equation which goes through (6, -5) and (-12, -5).x = -5A linear equation which is perpendicular to y = 12 and goes through (-5, 5).y = 5A linear equation which is parallel to y = 12 and goes through (-5, 5).y = -x + 5A linear equation (in slope-intercept form) which is perpendicular to y = x and goes through (3, 2).y = -5xA linear equation (in slope-intercept form) which goes through the origin and (1, -5).x = 2A linear equation which has undefined slope and goes through (2, 3).y = 3A linear equation which has a slope of 0 and goes through (2, 3).2x + y = -9A linear equation (in standard form) for a line with slope of -2 and goes through point (-1, -7).3x +2y = 1A linear equation (in standard form) for a line which is parallel to 3x + 2y = 10 and goes through (3, -4).y + 4 = 3/2 (x - 3)A linear equation (in point-slope form) for a line which is perpendicular to y = -2/3 x + 9 and goes through (3, -4).y - 8 = -0.2(x + 10)<span>The table represents a linear equation.
Which equation shows how (-10, 8) can be used to write the equation of this line in point-slope form?</span>

4 0

3 years ago

What is the speed (approximately) of a 2.5-kilogram mass after it has fallen freely from rest through a distance of 12 meters?

Norma-Jean [14]

Answer:

15.3 m/s

Step-by-step explanation:1) Find the gravitational potential energy using the equation :

$PE_g=mgh$