Posted by Megan on Monday, May 26, 2008 at 10:35am.
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.
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