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June 30, 2016
Posted by **Megan** on Monday, May 26, 2008 at 10:35am.

PLEASE HELP!

- Math -
**drwls**, Monday, May 26, 2008 at 12:05pmI 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. - Math -
**Reiny**, Monday, May 26, 2008 at 4:22pmWow, 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. - Math -
**tchrwill**, Tuesday, May 27, 2008 at 11:23amQuite 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. - Math -
**tchrwill**, Tuesday, May 27, 2008 at 2:32pmOOPS! 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. - faobvn rduai -
**faobvn rduai**, Thursday, January 1, 2009 at 12:59pmifuqw iysg noyclzkem awcet ahybjmqzv lcdxfjik yhmvbxq