Finding length of 3 sides of a triangle

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SUMMARY

The discussion focuses on calculating the lengths of the sides and the area of a triangle using trigonometric methods, specifically the cosine and Pythagorean theorem. The user initially miscalculated the lengths due to using radians instead of degrees, leading to confusion in the results. After correcting the angle mode, the final side lengths were determined to be BC = 32.6, AC = 20, and AB = 16.195, with the area calculated as 127.35 using the formula 0.5 * base * height.

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Homework Statement



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

Homework Equations



I have used both cosinus and pythagoras

The Attempt at a Solution



I already got that length of AC is 20 since that's 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 divide the triangle into 3 parts i get the length of AB to be 31 which is correct?

and my area is 0,5*height*b where the height 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:
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mimi.janson said:
CB i got to be 42,1 ...

How? What's your reasoning? It should be 32.6
 
dx said:
How? What's your reasoning? It should be 32.6

Because i first found the length of CD which i did with the sinus relation
c/sinC=d/sinD
c/sin23=10/sin67that 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? I've been staring at it for the whole day

i don't know if it plays any role but i use degrees instead of radians
 
mimi.janson said:
c/sin23=10/sin67that gave me the length of MD whish is 9,8.

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

DC = √(4.22 + 102) = 10.85

BC = 3DC = 32.6
 
Last edited:
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
 
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.
 
dx said:
MD = c = 10(sin(23)/sin(67)) = 4.2

DC = √(4.22 + 102) = 10.85

BC = 3DC = 32.6

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
 
mimi.janson said:
so my three lengths are now

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

Looks good.

mimi.janson said:
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
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:
dx said:
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.

ok i did so and i got 127,35
 
  • #10
Yes, it's correct.
 
  • #11
dx said:
Yes, it's correct.

Ok thank you so much you really made my day today <3
 
  • #12
No problem. Good luck with your exams.
 
  • #13
dx said:
No problem. Good luck with your exams.

thank you again and sorry i had to blabber about them i just get frustrated when doing math
 
  • #14
mimi.janson said:

Homework Statement



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
...

can anyone please check my attachment and help me out?
Please don't delete images from your posts.

It's quite frustrating for anyone willing to help you, to see that the initial information has suddenly disappeared.
 
  • #15
SammyS said:
Please don't delete images from your posts.

It's quite frustrating for anyone willing to help you, to see that the initial information has suddenly disappeared.

oh I am sorry must have clicked something without knowing, will be more carefull next time
 

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