Having three struts measuring 5.6m, 3.7m, 2.9m to form a triangular joist section, use appropriate formulae to calculate largest angle.

So by the numbers given I know this as an hypotonese triangle.

If I delegate each length a letter A B C I can get each angle by simply square rooting to letters? So for instance, a= square root of b*2 + C*2?

 

=Biggsy100;747439]Having three struts measuring 5.6m, 3.7m, 2.9m to form a triangular joist section, use appropriate formulae to calculate largest angle.

So by the numbers given I know this as an hypotonese triangle.

If I delegate each length a letter A B C I can get each angle by simply square rooting to letters? So for instance, a= square root of b*2 + C*2?

2.9^2 + 3.7^2 not equal to 5.6^2 m^2


Use sine/cosine rules.

 

can you please clarify what neq means please?

 

Sorry, I tried to use Latex editor but it failed. It was code, but I changed it.

 

Do you know sine/cosine rules?

 

So it seems a little long winded but I have designated side a to 2.9 metres: hence

2.9^2=3.7^2 +5.6^" - 2.3.7.56 Cos( α)

8.41 = 13.69 +31.36 - 41.44. Cos( α)

41.44. cos(α ) = 13.69 +31.36 -8.41

41.44. cos( α) = 36.64

cos (α) =0.88416988417

(a) = 27.85 degrees?

*** This is the part you tell me it can be done in under 5 seconds

 

So it seems a little long winded but I have designated side a to 2.9 metres: hence

2.9^2=3.7^2 +5.6^" - 2.3.7.56 Cos( α)

8.41 = 13.69 +31.36 - 41.44. Cos( α)

41.44. cos(α ) = 13.69 +31.36 -8.41

41.44. cos( α) = 36.64

cos (α) =0.88416988417

(a) = 27.85 degrees?

*** This is the part you tell me it can be done in under 5 seconds

Yep, you got it. Five seconds is a bit of an over statement, but it speeds up with practice.

Maths is fun!

 

@Biggsy100

I knew that angle seemed tiny, I wasn't paying attention. The largest angle is opposite the longest side.

You need to recalculate!

 

2.9^2=3.7^2 +5.6^" - 2.3.7.56 Cos( α)

I am lost at the end of this equation. 5.6 is raised to what? And what does 2.3.7.56 represent? It's not a number…

 

I will have to work on this and give some more thought as I am not sure how to get the angle with just the strut measuremenst

 

The Cosine Rule: In any triangle the square of one side equals the sum of the squares of the other 2 sides minus twice the product of the same two sides and the cosine of their included angle:
a^2 = b^2 + c^2 - 2bc Cos A
or Cos A = b^2 + c^2 - a^2 / 2bc
and Cos B = a^2 + c^2 - b^2 / 2ac

 

When you need help always remember the old Indian tribe, SOHCAHTOA. Have fun.

 

Is this better?

a^2 = b^2 + c^2 - 2.b.c cos(a)

Substituting a = 5.6, b = 3.7,c=2.9

5.6^2= 3.7^2 +2.9^2-2.37.2.9 cos(a)

31.36 = 13.69 +8.41 -21.46. cos (a)

21,46= cos(a) =13.69+8.41-31.36

21.46.cos (a) = -9.26

cos(a) = - 0.431

a = arcoss (0.431)

(a) = 115.56

 

That's what I get Biggsy - SOHCAHTOA, I was told it was a mountain in Japan with sides of ratio 3,4,5. It would have been easier if our maths teacher, at school, had told us this, I didn't come across it till later. I can remember the Cosine Rule in words rather than the equation.

 

SOH-CAH-TOA does not apply to an obtuse triangle. It holds for right-angled triangles.

 

SOH-CAH-TOA does not apply to an obtuse triangle. It holds for right-angled triangles.

Have I got it correct this time?

 

Yep, largest angle is opposite the longest side.

 

b^2= a^2 + C^2 - 2.a.c Cos(B)

After substitution b = 3.7, a =5.6, c= 2.9

b^2=5.6^2 + 2.9^2 - 2. 5.6 .2.9 Cos B

13.69 = 31.36 +8.41 - 32.48. Cos B

32.48. Cos(B) = 31.36 +8.41 - 13.69

32.48 Cos(B) = 26.08

Cos (B) = 0.802

(B) = arcos (0.802)

(B) = 36.58

 

Are you trying to solve for all angles?

Yep it sure is, but try substituting your numbers at the very end. It's less error prone and you can see what you're looking at before you make the substitution.

b^2 = a^2 + c^2 - 2ac cos(B)

2ac cos(B) = a^2 + c^2 - b^2

Angle B = Arc cos [(a^2 + c^2 - b^2)\(2ac)]

Angle B = Arc cos [(5.6^2 + 2.9^2 - 3.7^2)\(2)(5.6)(2.9)]

Angle B = 36° 35' 12.19"

Optional method of expressing the angle using the base-60 Sexagesimal system, the (° ' ") on your calculator. It's an exact plane angle and you can retrieve it in pure decimal format if you wish.

Angle B = 36 + (35\60) + (12.19\60^2) = 36.58671944° ≈ 36.59°

 

Thank you for that! Should I apply the same methods to a base plate calculation? I know I need to split the calculation or visualise the base plate split into two?

The problem I have is:

A design brief indicates three holes A,B,C to be drilled in a plate, with a=73mm, b=117mm and a =37 degrees

I need to find the possible value for the dimension of C? Should I simply split into 2 triangles and measure accordingly?