"The diagram shows the roof of a building. The base ABCD is..."

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KaivettuJose

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Can someone please help me solve question 3b shown on the image

Maths question 2.JPG

The answer is 75.5 but I don't know how to solve it.
Any help is appreciated.
 
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KaivettuJose said:
Can someone please help me solve question 3b shown on the image

The diagram shows the roof of a building. The base ABCD is a horizontal rectangle 7m by 4m. The triangular ends are equilateral triangles. Each side of the roof is an isosceles trapezium. The length of the top of the roof, EF, is 5m. Calculate... (b) the side of angle EBC.

The answer is 75.5 but I don't know how to solve it.
Any help is appreciated.
Click to expand...

What recent rules/theorems/formulas have they covered (in particular, in the section that generated this homework question)?

When you reply, please include a clear listing of your thoughts and efforts so far. (Start with the assumption that the ridgeline is centered, and then the "altitude" line that you dropped from E to BC to form a right triangle.) Thank you!
 
@KaivettuJose

Are you another one we're never going to hear from again?

Have you managed to figure out how the correct answer (75.5°) is arrived at yet?

If so it would have been courteous of you to let us know (please do so) but, if not, below is a more detailed "walkthrough" that will help you (or anyone else viewing this thread that isn't sure how to proceed given the help already provided) to reach the correct solution.

Here is the drawing (that you should have sketched yourself at @Dr.Peterson's suggestion) viewing the side of the roof when you are looking "straight on" to it.
roof.png​
Please make your own sketch of this (or copy mine) and then you can follow the procedure (outlined below) to replace the coloured question marks and calculate the measure of ∠B.

The fact that it is an isosceles trapezium means that BC & EF are parallel, it is symmetrical (about a mid-line through BC & EF) and FC is the same length as BE (ie: 4 m).

You know that EF is 5 m and BC is 7 m.

What @Dr.Peterson suggested is that you should drop perpendiculars from E to G and from F to H (as shown by the red dotted lines).

Now, EFHG is a rectangle.
So how long is GH? (1st "?" to replace.)

And then (due to the symmetry) how long will BG and HC be? (Remaining two "?"s to replace.)

Now consider the right-angled triangle EBG.
You are given the length of its hypotenuse (4 m) and you now know the length of the side adjacent to ∠B.

Using basic Right-Angled Triangle Trigonometry (SOH-CAH-TOA?) it should now be a simple matter for you to calculate the measure of ∠B. Yes?

If you were struggling with part b) of this question, I suspect you might have had further difficulty working on part c) too but working your way through part b) in the fashion outlined above should (I trust) help you to tackle part c) as well.

We would be grateful if you would post back a similar sketch to the one above with the question marks replaced by the correct values and a calculation of the measure of ∠B shown.

Thank you. ?
 
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