One more question.?

Question

One last trig question, thanks to everyone who's helped with my questions.

First person to answer this completely gets best answer.

A triangle, ABC, has vertices A(-1, 1) B(5,1) and C (2, 4).

Determine perimeter of ABC and express as an exact value in simplest radical form.

Determine the slope of AC.

Find the measure of the angle A.

What is the area of the triangle ABC?

Answer 1

AB = 6 = base of triangle
height of triangle = 3

area of triangle ABC = .5(6)(3) = 9

slope of AC = (4 - 1) / (2 - (-1)) = 3 / 3 = 1

ABC is an isosceles triangle; AC and BC are equal.

AC = BC = 3 sqrt 2

perimeter(ABC = 3 + 6 sqrt 2

measure of angle A = 45 degrees

So, to recap:

P = 3 + 6 sqrt 2
AC has slope of 1
measure of A = 45
area = 9 units

Answer 2

1) Perimeter

Find the distance between each of the three points to get the length of each side of the triangle. Then add the sides up to get perimeter.

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

AB = d1 = sqrt[ (1 - 1)^2 + (5 - (-1))^2 ] = sqrt[ 0 + 36 ] = 6
BC = d2 = sqrt[ (4 - 1)^2 + (2 - 5))^2 ] = sqrt [ 9 + 9 ]
= sqrt[18] = 3*sqrt(2)
AC = d3 = sqrt[ (1 - 4)^2 + (-1 - 2)^2 ] = sqrt[ 9 + 9 ]
= sqrt[18] = 3*sqrt(2)

Perimeter = d1 + d2 + d3 = 6 + 3*sqrt(2) + 3*sqrt(2)
= 6 + 6sqrt(2) = 6[1 + sqrt(2) ]

2) Slope of AC

Slope = (y2 - y1) / (x2 - x1) = (4 - 1) / (2 - (-1)) = 3/3 = 1


3) Angle A is the angle between sides B and C

Use the cosine law

a^2 = b^2 + c^2 - 2bc cosine(angle between sides B and C)

or in this example:

BC^2 = AB^2 + AC^2 - 2*AB*AC*cosine(angle A)
[3*sqrt(2)]^2 = [3*sqrt(2)]^2 + [6]^2 - [2*3*sqrt(2)*6*cosine (A)]
18 = 18 + 36 - [36*sqrt(2) * cosine (A)]
[-36 / -36*sqrt(2)] = cosine (A)
cosine (A) = 1/sqrt(2)

Use a calculator inverse cosine button or use the standard triangle ratio for a 45-45-90, which is 1-1-sqrt(2) and the hypotenuse is sqrt(2), the two legs are 1 and 1. Since cosine (A) is adjacent/hypotenuse, adjecent = 1 and hypotenuse = sqrt(2), so, A = 45 degrees.

4)Area

The general formula for the area of any triangle is:

A = sqrt[ (s)(s - AB)(s - BC)(s - AC) ] with s = (AB + BC + AC)/2

First find "s" s = (6 + 3sqrt(2) + 3sqrt(2) ) / 2
=(6 + 6sqrt(2) ) / 2 = 3 + 3sqrt(2))

Area = sqrt[ (3 + 3sqrt(2))(3 + 3sqrt(2) - 6)(3 + 3sqrt(2) - 3sqrt(2))(3 + 3sqrt(2) - 3sqrt(2)) ]
=sqrt [ (3 + 3sqrt(2))(3sqrt(2) - 3)(3)(3) ]
=sqrt [ (9sqrt(2) -9 + 18 - 9sqrt(2)(9) ]
=sqrt [ (9)(9) ] = 9

Answer 3

vectors
AB = (6,0) length 6
BC = (-3, 3) length 3V2
CA = (-3,-3) length 3V2

The perimeter is AB+BC+CA = 6 + 6V2

Slope of AC is (-3)/(-3) = 1

To find angle, determine dot product of AB and AC:

AB . AC = (6,0) . (3,3) = 18

divide by the products of the length of AB and AC:

18 / (6 . 3V2) = 1/V2

This is the cosine of 45 degrees.

The triangle is isosceles, with base AB = 6 and height 3; the area is (3 . 6) / 2 = 9

Answer 4

Summer school sucks doesnt it

Answer 5

who do u think we are albert einstin

Answer 6

ask your math teacher.