• Sep 19th 2005, 04:22 AM
emydawn
The three points A, B, and C, with coordinates A(0,1), B(3,4), and C(1, 3) respectively, are joined to form a triangle:

a) Show that COS ACB= -4/5

b) Calculate the area of triangle ABC.

Ive done this question and have come out with the answer a) 8.66 but apparently its 5. ??
• Sep 19th 2005, 05:50 AM
ticbol
Here is one way.

Since the coordinates of the corners are given, then we can get the lengthjs of the 3 sides
distance = sqrt[(x2 -x1)^2 +(y2 -y1)^2]

Then we can get the cos (angle ACB) by using the Law of Cosines.
(AB)^2 = (AC)^2 +(BC)^2 -2(AC)(BC)*cos(angle ACB)
cos(angle ACB) = [(AC)^2 +(BC)^2 -(AB)^2] / [2(AC)(BC)] ---(1)

Then we can get the area of triangle ABC by using the Heron's Formula.
A = sqrt[s(s-a)(s-b)(s-c)]
Or, by using the formula, A = (1/2)bc*sinA

Those above need long computations. I don't need to show those computations here, so solve them yourself.

I solved them on paper here and I got:
AB = sqrt(18)
BC = sqrt(5)
AC = sqrt(5)

Substituting those in (1), I got cos(angle ACB) = -8/10 = -4/5.

Then using the formula A = (1/2)bc*sinA for the area of triangle A,
where
angle "A" = angle ACB
"b" = AC = sqrt(5)
"c" = BC = sqrt(5)

cos(A) = -4/5
Square both sides,
cos^2(A) = 16/25
So,
1 -sin^2(A) = 16/25
sin^2(A) = 1 -16/25 = 9/25
sinA = sqrt(9/25) = 3/5

So, area of triangle ABC
= (1/2)*sqrt(5)*sqrt(5)*(3/5)
= (1/2)(5)(3/5)
= 3/2 or 1.5 sq. units ----------answer.
• Sep 19th 2005, 12:12 PM
emydawn
haha i just went to write it down and do the working myself then i realised i have actually done this. i typed out the wrong question godammit :o

the question i meant to type was:

In triangle ABC, AB= 10cm, BC= a√3 cm, AC=5√13 cm and angle ABC=150 degrees.
Calculate:

a) the value of a

b) the exact area of triangle ABC

this is the question that i got 8.66 for, but the textbook answer says it is 5.

howww?
thankss

xxx
• Sep 20th 2005, 03:04 AM
ticbol
I don't know, but this looked like a trap. Not so good. But I'd give you the benefit of the doubt. Let us say you really made a mistake.

--------------
Same thing:
>>>Law of Cosines.
>>> Area = (1/2)ab*sinA

[5sqrt(13)]^2 = (10)^2 +[a*sqrt(3)]^2 -2(10)[a*sqrt(3)]*cos(150deg)
25(13) = 100 +3a^2 +30a
325 = 3a^2 +30a +100
0 = 3a^2 +30a +100 -325
0 = 3a^2 +30a -225
Divide both sides by 3,
0 = a^2 +10a -75
a^2 +10a -75 = 0
(a+15)(a-5) = 0
a = -15 or 5 ------------***
Since there are no negative dimensions, reject a = -15.