The sides of a triangle form an arithmetic sequence with a common difference of 2. The ratio of the measure of the largest angle to that of the smallest angle is 2:1. Find the area of the triangle.
PLEASE HELP!
I found by using the "triangle solver" at
(Broken Link Removed)
, and by trying a series of integer side lengths starting with 3:5:7, 4:6: 8 etc. that the side length ratio of the triangle is exactly 8:10:12.
The area of such a triangle is sqrt 1575 = 15 sqrt 7 = 39.686 square units.
Since I figured that an iterative solution was necessary, I feel justified doing it this way. Your teacher may not think so, howwver.
Wow, good question! I tried it this way
let the smallest side be x, and the angle across from it ß
then sin(2ß)/(x+4) = sinß/x
2sinßcosß/(x+4) = sinß/x
dividing by sinß and solving for cosß
cosß = (x+4)/(2x)
I then had no other choice but to use drwls method of iteration, starting with 3,5,7 since any smaller numbers don't produce a triangle
I neede an integer value of x so I could find ß from cosß.
I then had to test if sinß/x is equal to sin(2ß)(x+4) and that happened when x=8
Once I had x=8, ß was 41.41º and the area
= 1/2(10)(12)sin41.41º getting the same answer as drwls.
Quite a challenge.
Let the three angles be A, B and C and the three opposite sides be a, b and c.
1--C = 2A
2--sinA/a = sinC/c = sinC/(a+4)
3--sinC/sinA = (a+4)/a
4--sin2A/sinA = (a+4)/a
5--2sinAcosA/sinA = (a+4)/a
6--cosA = (a+4)/2a
7--sinA/a = sinB/b
8--sinA = sinB/(a+2)
9--sinB/sinA = (a+2)/a
10--sin(180-3A)/sinA = (a+2)/a
11--[sin(180)cos3A - cos(180)sin3A]/sinA = (a+2)/a
12--sin3A/sinA = (a+2)/a
13--[3sinA - 4sin^3A]/sinA = (a+2)/a
14--(3 - 4sin^2A) = (a+2)/a
15--4sin^2A = 3 - (a+2)/a leading to sin^A = (2a-2)/4a
16--From sin^A + cos^A = 1, (2a-2)/4a + (a+4)^2/4a^2 = 1
17--Multiplying and simplifying, a^2 - 6a - 16 = 0
18--Therefore, a = [6+/-sqrt(36 + 64)]/2 = [6+/-10]/2 = 16/2 = 8.
19--Ths, the three sides are 8, 10 and 12.
OOPS! Forgot to compute the area.
The sides of a triangle form an arithmetic sequencee witrh a common difference of 2. The ratio of the measure of the largest angle to that of the smallest angle is 2:1. Find the area of the triangle.
Let the three angles be A, B and C and the three opposite sides be a, b and c.
1--C = 2A
2--sinA/a = sinC/c = sinC/(a+4)
3--sinC/sinA = (a+4)/a
4--From (1) sin2A/sinA = (a+4)/a
5--2sinAcosA/sinA = (a+4)/a
6--cosA = (a+4)/2a
7--sinA/a = sinB/b
8--sinA = sinB/(a+2)
9--sinB/sinA = (a+2)/a
10--sin(180-3A)/sinA = (a+2)/a
11--[sin(180)cos3A - cos(180)sin3A]/sinA = (a+2)/a
12--sin3A/sinA = (a+2)/a
13--[3sinA - 4sin^3A]/sinA = (a+2)/a
14--(3 - 4sin^2A) = (a+2)/a
15--4sin^2A = 3 - (a+2)/a leading to sin^A = (2a-2)/4a
16--From sin^A + cos^A = 1, (2a-2)/4a + (a+4)^2/4a^2 = 1
17--Multiplying and simplifying, a^2 - 6a - 16 = 0
18--Therefore, a = [6+/-sqrt(36 + 64)]/2 = [6+/-10]/2 = 16/2 = 8.
19--Thus, the three sides are 8, 10 and 12.
20--cosA = (8+4)/2(8) = 12/16 making A = 41.40962 deg. and C = 82.8192 deg.
21--From Heron's area formula, the area A = sqrt[30(18)20(22] = 39.686 sq. units.
22-- Area check: A = 8(10)sin82.8192)/2 = 39.686 sq. units.
To find the area of a triangle, we need to know the lengths of its sides. In this case, we know that the sides of the triangle form an arithmetic sequence with a common difference of 2. Let's call the smallest side "a," the middle side "a + 2," and the largest side "a + 4."
Now, let's use the fact that the ratio of the measure of the largest angle to the smallest angle is 2:1. The angles of a triangle are related to the lengths of its sides using the Law of Cosines. According to the Law of Cosines, the measure of the largest angle, opposite the largest side, can be found using the formula:
cos(A) = (b^2 + c^2 - a^2) / (2bc),
where "a," "b," and "c" are the lengths of the sides of the triangle, and A is the angle opposite side "a."
Using this formula, we can find the measure of the largest angle, A:
cos(A) = [(a + 4)^2 + (a + 2)^2 - a^2] / [2(a + 4)(a + 2)].
Next, we need to find the measure of the smallest angle, B. The smallest angle is opposite the smallest side "a." Since the sides are in an arithmetic sequence, we can use the Law of Cosines again, substituting "a + 4" for "a" to find the measure of angle B:
cos(B) = [(a + 2)^2 + a^2 - (a + 4)^2] / [2(a + 2)a].
We also know that the sum of the three angles in a triangle is 180 degrees. So, we can find the measure of the middle angle, C, by subtracting the measures of angles A and B from 180 degrees:
C = 180 - A - B.
Now, we have the measures of all three angles in terms of "a.". Once we find the values of A, B, and C, we can use the formula for the area of a triangle:
Area = ½ * base * height.
To find the base and height, we need to determine which side of the triangle is the base. Do you have any additional information about the triangle?