Finding length of 3 sides of a triangle

1. Jan 6, 2013

mimi.janson

1. The problem statement, all variables and given/known data

hi i have a photo with some given information. I need to find the length of all three sides of that triangle and also the area of the triangle

I have already found them but i get 2 different results

2. Relevant equations

I have used both cosinus and pythagoras

3. The attempt at a solution

I already got that length of AC is 20 since thats one of the given lengths. CB i got to be 42,1 and AB i sometimes get to be 19,2 when i use cosiunus, but when i use pythagoras and devide the triangle into 3 parts i get the length of AB to be 31 which is correct?

and my area is 0,5*hight*b where the hight is 3*the length of MD. so i got MD to be 9,89 and therefore my area is 0,5*29,67*20 = 296 but it just seems too big?

can anyone please check my attachment and help me out?

Last edited: Jan 6, 2013
2. Jan 6, 2013

dx

How? What's your reasoning? It should be 32.6

3. Jan 6, 2013

mimi.janson

Because i first found the length of CD which i did with the sinus relation
c/sinC=d/sinD
c/sin23=10/sin67

that gave me the length of MD whish is 9,8.
since i had the length of MD =9,8 and the length of MC =10 i used pythagoras 102*9,82=the length of DC.

so i got DC to be about 14 and since BC is 3 times DC i just had to say 3*14=42

did i really solve it wrong? ive been staring at it for the whole day

i dont know if it plays any role but i use degrees instead of radians

4. Jan 6, 2013

dx

MD = c = 10(sin(23)/sin(67)) = 4.2

DC = √(4.22 + 102) = 10.85

BC = 3DC = 32.6

Last edited: Jan 6, 2013
5. Jan 6, 2013

dx

A simpler way to calculate DC is from the definition of cosine:

DC cos(23) = MC, so DC = MC/cos(23) = 10/cos(23) = 10.86

6. Jan 6, 2013

dx

I see where you went wrong. When you calculated sin(23) and sin(67) on your calculator, you were in radian mode, and the calculator interpreted 23 as 23 radians, instead of 23 degrees. You should switch to degree mode.

7. Jan 6, 2013

mimi.janson

Yes i checked and it was the degrees that were the problem...im really nervous for my exams this summer, because you see i have been looking at this which should be a simple matter since they taught it to my class like 2 years ago, and i still spent like 4 hours trying to solve it because of the radians...and i never find the fail ever even though i have tons of this kind of strange fails....sometimes i even see a number wrong and sit trying to find it for hours. .. but i have changed it now and do get the results you get too.

I prefer to use pythagoras instead of cosine as much as possible because i find it hard.
I used cosine to find the length AB and got it to be 16,195

so my three lengths are now

BC=32,597
AC=20
AB = 16,195

i was just thinking does that make an area of 127,35 realistic? i said 0,5*12,735*20. I used the length by saying that it must me 3*MD

8. Jan 6, 2013

dx

Looks good.

0.5*12.735*30 (not 20) is the area of the right triangle with hypotenuse BC. From that you have to subtract the area of the right triangle with hypotenuse BA.

Last edited: Jan 6, 2013
9. Jan 6, 2013

mimi.janson

ok i did so and i got 127,35

10. Jan 6, 2013

dx

Yes, it's correct.

11. Jan 6, 2013

mimi.janson

Ok thank you so much you really made my day today <3

12. Jan 6, 2013

dx

No problem. Good luck with your exams.

13. Jan 6, 2013

mimi.janson

thank you again and sorry i had to blabber about them i just get frustrated when doing math

14. Jan 6, 2013

SammyS

Staff Emeritus