Calculating Length of Cottage Rafters

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SUMMARY

The discussion centers on calculating the length of cottage rafters for a design that is 15 meters wide, with rafters meeting at an 80-degree angle and overhanging the supporting wall by 0.5 meters. Participants clarify that the problem can be solved using trigonometry, specifically right triangles, rather than the sine or cosine laws. By constructing a diagram, the base of the isosceles triangle is determined to be 16 meters (15 m width plus 2 x 0.5 m overhang), and the angle at the top is 40 degrees. The solution involves using the sine function to find the hypotenuse, which represents the length of the rafters.

PREREQUISITES
  • Understanding of basic trigonometry concepts, particularly right triangles
  • Familiarity with the sine function and its application in solving triangles
  • Ability to interpret geometric diagrams and apply them to real-world problems
  • Knowledge of isosceles triangles and their properties
NEXT STEPS
  • Study the application of the sine function in right triangles
  • Learn how to construct and interpret geometric diagrams for trigonometric problems
  • Explore the properties of isosceles triangles and their relevance in construction
  • Practice solving real-world problems involving angles and lengths using trigonometry
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Students studying geometry and trigonometry, architects and builders involved in cottage design, and anyone interested in applying mathematical principles to construction projects.

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Kelly designs a cottage that is 15 m wide. The roof rafters are the same length and meet at angle of 80 degrees. The rafters hang over the supporting wall by 0.5 m. How long are the rafters?

Im not sure how to set up this question will it involve algebra?

I don't know how the cosine law or sine law can be used can someone help me out please? :smile:
 
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It requires, of course, trigonometry. You don't need the sine and cosine laws since this can be done entirely with right triangles.

First draw a picture! You have an isosceles triangle with the roof forming the two top (equal) lines. Since "the rafters hang over the supporting wall by 0.5 m" (I am assuming that that is measured horizontally) the length of the base of that triangle is 15 m (the 15 m width plus the two 0.5 m overhang). You can get right triangles by drawing the vertical line down from the crest of the roof (i.e. the top angle). That way you have two identical right triangles. The angle at the top is (1/2)(80)= 40 degrees and the length of the "opposite side" is (1/2)(16)= 8 m. You know that
sin(angle)= opposite/hypotenuse so sin(40)= 8/x. Of course, the rafters ARE the hypotenuses.
 
I think my diagram doesn't look right because I don't understand why you added the 0.5 to the base. why wasn't it added to the rafters?

Can someone please explain? :confused:
 
Last edited:
aisha said:
Kelly designs a cottage that is 15 m wide. The roof rafters are the same length and meet at angle of 80 degrees. The rafters hang over the supporting wall by 0.5 m. How long are the rafters?

Im not sure how to set up this question will it involve algebra?

I don't know how the cosine law or sine law can be used can someone help me out please? :smile:
What I do not get is how are the rafters arranged. Is there a diagram with it or are you simply given the question as it is written?? Also, why does the question say that the rafters are the same length and then say they are 0.5m longer??

Answer these and the answer might be obvious.

The Bob (2004 ©)
 
aisha said:
I think my diagram doesn't look right because I don't understand why you added the 0.5 to the base. why wasn't it added to the rafters?

Can someone please explain? :confused:

That possibility had occurred to me- you might want to ask your teacher to clarify it. I decided that the words "hang over" referred to the distance out from the wall.

If you think it means that the length of the rafter, past the wall, is 0.5 m, do the problem with base 15 m so the "opposite side" of the right triangle is 7.5 m (the fact that the other way gives an integer length here may have influenced me!). Now solve for the hypotenuse of that right triangle and then add 0.5 m to it.
 
ok thanks soo much, I will ask her for sure! :smile:

Bob there was no diagram just the question I didnt write it so I don't know why it says what it does. :smile:
 
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