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. ??

2. 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.

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

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

4. 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.

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Area = (1/2)(10)[a*sqrt(3)]*sin(150deg)
Area = (1/2)(10)[5sqrt(3)]*sin(150deg)
Area = (1/2)(10)[5sqrt(3)](0.5)

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### maths formula ab=10cm bc=10 cm ac=?

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