In the triangle ABC given, side AB is twice as long as side AC. We have:
m<BAC=120°
mBC= 8cm
Determine rounded to the nearest unit, the perimeter of triangle ABC.
Let AC = x
let AB = 2x
We can use the cosine law to find x
x^2 + (2x)^2 - 2(x)(2x)cos120° = 8^2
5x^2 - 4x^2(-1/2) = 64
7x^2 = 64
x^2 = 64/7
x = 8/√7
perimeter = x+2x+8
= 3x + 8
= 24/√7 + 8 = appr 17
To find the perimeter of triangle ABC, we need to use the information given and apply the properties of triangles.
First, we'll refer to the fact that side AB is twice as long as side AC. Let's assume the length of side AC is x. Then, the length of side AB becomes 2x.
Next, we'll consider the given angle measure of ∠BAC, which is 120°. This angle is opposite side BC in the triangle.
To find the length of side BC, we can use the Law of Cosines, which states that for any triangle with sides a, b, and c, and angle C opposite side c, we have:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, we know side AC (a), side AB (b), and the angle ∠BAC (C). So, we can rearrange the formula to solve for c:
BC^2 = AC^2 + AB^2 - 2 * AC * AB * cos(∠BAC)
Now, plugging in the known values:
BC^2 = x^2 + (2x)^2 - 2 * x * 2x * cos(120°)
BC^2 = x^2 + 4x^2 - 4x^2 * (-0.5)
BC^2 = x^2 + 4x^2 + 2x^2
BC^2 = 7x^2
Since BC is given as 8 cm, we can solve for x:
7x^2 = 8^2
Simplifying,
49x^2 = 64
x^2 = 64/49
x ≈ 1.1836
Now, we can find the length of side AB:
AB = 2x
AB ≈ 2 * 1.1836
AB ≈ 2.3672
To find the perimeter, we add up the lengths of all three sides:
Perimeter = AB + AC + BC
Perimeter ≈ 2.3672 + 1.1836 + 8
Perimeter ≈ 11.5508
Rounded to the nearest unit, the perimeter of triangle ABC is approximately 12 cm.