Trig question...?

Question

The three given points are the vertices of a triangle. Solve this triangle rounding lengths of sides to the nearest tenth and angle measures to the nearest degree.

A (0,0), B (4, -3), C (1, -5)

Answer 1

AB=sqrt((xa-xb)^2+(ya-yb)^2)=sqrt(16+9)=5
AC=sqrt((xa-xc)^2+(ya-yc)^2)=sqrt(1+25)=sqrt(26)~5
BC=sqrt((xb-xc)^2+(yb-yc)^2)=sqrt(9+4)=sqrt(13)~4
Let be S=area of triangle ABC => S~9. We have;
2S=AB*AC*sinA=AB*BC*sinB=AC*BC*sinC then we have:
sinA=2S/AB*AC=18/25=0.72 =>A=46 degree rounding
sinB=2S/AB*BC=18/20=0.9 =>B=65 degree rounding
sinC=2S/AC*BC=18/20=0.9 =>C=64 degree rounding

Answer 2

I try to help
length of AB . It is the hypothenuse of a triangle rectangle that you can imagine when you draw the perpendicular to the x abciss (I regret not can draw) the sides are 4 and 3 and the hypothenuse is
H^2 = 4^2+3^2 = 25 so H =AB = 5
For AC same way you draw a projection onthe x axis and you have
AC^2 = 25+1 ; AC = 26^0.5= 5.1
for BC you can also imagine a rectangle with length of the sides differences of ordinates which is 2 (-3 -(-5))and differences of abcisses which is 3 so BC^2 = 3^2+2^2 =13 and BC=3.6

so the tree sides are AB =5, AC =5.1 , BC =3.6

from sinus relation if you call A angle at vertex A , B angle vertex B, C angle vertexC

you have 3.6/sinA = 5.1/sinB = 5/sinC

No idea more

Answer 3

The first step in this problem is to determine the length of AB, BC, and AC. Since these are points on a graph, the sides of the length is best done using the distance formula.

A reminder that the distance formula goes as follows:

d = sqrt ( [x2 - x1]^2 + [y2 - y1]^2 )

AB = sqrt ( [0 - 4]^2 + [0 - (-3)]^2) = sqrt(16 + 9) = sqrt(25) = 5
AC = sqrt ( [0 - (-5)]^2 + [0 - 1]^2) = sqrt(25+1) = sqrt(26)
BC = sqrt ( [4 - 1]^2 + [-3 - (-5)]^2 = sqrt(9+4) = sqrt(13)

Let's call the angles A, B, and C as the angles at the points A, B, and C. This means that we have to assign sides a, b, and c.
a = side BC = sqrt(13)
b = side AC = sqrt(26)
c = side BC = 5

The cosine formula goes as follows

a^2 = b^2 + c^2 - 2bc(cosA)

Since we want angle A, all we have to do is solve for cosA

a^2 - b^2 - c^2 = -2bc(cosA)

Multiply both sides by -1 to get

-a^2 + b^2 + c^2 = 2bc(cosA)

Divide both sides by 2bc, to get

cosA = [-a^2 + b^2 + c^2]/(2bc), so
A = cos(inverse) { [-a^2 + b^2 + c^2]/(2bc) }

Similarly,

B = cos(inverse) { [-b^2 + c^2 + a^2]/(2ac) }
C = cos(inverse) { [-c^2 + a^2 + b^2]/(2ab) }

So it's just a matter of plugging in your points.
a = sqrt(13)
b = sqrt(26)
c = 5

A = cos(inverse) { [-13 + 26 + 25]/[2*sqrt(26)*5] }
A = cos(inverse) { [38]/[10sqrt(26)] }
A = cos(inverse) {19/[5sqrt(26)]

B = cos(inverse) { [-26 + 25 + 13]/(2*sqrt(13)*5) }
B = cos(inverse) {12/(10sqrt(13)) }
B = cos(inverse) {6/[5sqrt(13)] }

C = cos(inverse) { [-25 + 13 + 26]/(2*sqrt(13)*sqrt(26)) }
C = cos(inverse) { 14/[2*13*sqrt(2)] }
C = cos(inverse) { 7/[13*sqrt(2)] }

I'll leave it up to you to enter those values into your calculator. Remember your calculator has to be in RAD mode.

Answer 4

Graphing it out is the easiest way to see it.....it's not a right triangle by itself, so use each side of the triangle as the hypotenuse of a right triangle using the axes of the graph and imaginary lines to supplement. Your 3 side lengths become 5, 3.6, and 5.1.......you can figure out the angles using the same method of 3 seperate right triangles and using standard trig.

Answer 5

This is a vector question
First find the length or the vectors (sqrt(x^2 + y^2))
then use the dot product rule ie A.B = cos(angle)

Answer 6

i got here upon trig to be a lot less annoying. Geometry in intense college replaced into the hardest for me, in reality, I nevertheless imagine intense college geometry is fairly harder than a good type of the faculty math instructions i'm taking to be uncomplicated...I disliked it very a lot.

Answer 7

are we allowed to do other peoples homework on here? don't be a lazy SOB. get your notes out.