GCSE Q worth 4 marks: The diagram shows two vertical masts A and B....

Kindly please re-read the "Read Before Posting" message, and reply with the requested information, starting with the height that you obtained for Mast B (hint: multiply) and the right triangle that you drew.

Thank you!

Eliz.
 
davonovo said:
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In any question like this it is always advisable to begin with a sketch.
NB: A sketch is not (necessarily) an "accurate" diagram.

So, you should start with a sketch similar to how I have modified the 'original' (below, qv) but replacing the question marks with the correct values.

Once you have done that it is a simple matter of Right-Angled Triangle Trigonometry to calculate the length of the green line.

Masts.jpg​

Please now come back showing us your sketch and your calculation of the distance between the two masts.

I trust we will hear back from you soon. ?

Hope that helps? ?
 
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PS: All you really need in your sketch is the red, blue & black, right-angled triangle (which you might label at the vertices as: XYZ with (corresponding) sides: x, y & z) with the two "given" dimensions shown on it to support your calculation of the length of its base but a fuller sketch (more like mine) would also illustrate how you arrived at those dimensions.
 
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Thank you all. I realise my mistake - I read the angle of depression as the angle inside the top of the triangle, instead of 65, so my trigonometry using tan was wrong. Thanks for the responses
 
It's all very well Like-ing the work I have done (for you) but we want to see your work now!

Please show us your sketch and calculation(s).
 
davonovo said:
Here is my solution. I think I have it right now. Thanks
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Yes, that is (almost) correct.

I don't understand why you felt the need to rotate the triangle though there is nothing "wrong" in doing so, however, as I have pointed out (repeatedly) to some other member(s) 105 × tan 65° ≠ 225.17!

You should write: 105 × tan 65° ≈ 225.17 or 105 × tan 65° = 225.17 to 2 d.p. (or 5 s.f.)
​
Your work might also be criticized for the fact that (the brevity of) your sketch, although it provides sufficient information to support your final calculation, it does not show where you got the 105 m or the 65° from! (Your "calculations" of those values just appear "out of the blue".) ?

I did point that out in my earlier post but it is, perhaps, a minor criticism (the 'unqualified rounding' of your answer isn't).
 
The Highlander said:
Yes, that is (almost) correct.

I don't understand why you felt the need to rotate the triangle though there is nothing "wrong" in doing so, however, as I have pointed out (repeatedly) to some other member(s) 105 × tan 65° ≠ 225.17!

You should write: 105 × tan 65° ≈ 225.17 or 105 × tan 65° = 225.17 to 2 d.p. (or 5 s.f.)
​
Your work might also be criticized for the fact that (the brevity of) your sketch, although it provides sufficient information to support your final calculation, it does not show where you got the 105 m or the 65° from! (Your "calculations" of those values just appear "out of the blue".) ?

I did point that out in my earlier post but it is, perhaps, a minor criticism (the 'unqualified rounding' of your answer isn't).
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To be fair, the poster's work may have omitted material that came before the work shown, which is the last part; the poster's work is fairly well correct.
 
davonovo said:
Here is my solution. I think I have it right now. Thanks
View attachment 35917
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To davonovo, your work work is correct. I is good that you worked it out for yourself.
This in a quibble at all, but simply to give you another way to look at the question.
The two dashed red lines indicating the horizonal are parallel.
Therefore a line from the top of pole B to the top of pole A is a transversal.
It creates congruent alternate interior angles.
Thus angle in the lower-left-hand of the blue-red-gray triangle measures [imath]25^{o}[/imath]
You found the difference in highs is [imath]105m[/imath] using [imath]x[/imath] as the distance between poles
we see [imath]\tan(25^o)=\dfrac{105}{x}[/imath] or [imath]x=\dfrac{105}{\tan(25^o)}[/imath]

[imath][/imath][imath][/imath]
 
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