trig help (1 Viewer)
- Thread starter yashbb
- Start date
I suggest having a look at the following thread, where the same question was asked, with only a few values being different (angles of elevation= 42° and 35° instead of 36° and 27° respectively and PQ = 200 instead of 250). Essentially, the working out is the same as long as you use the respective values from your question:
I hope this helps!
I hope this helps!
Thank you it does help, but could you please explain part b again?I suggest having a look at the following thread, where the same question was asked, with only a few values being different (angles of elevation= 42° and 35° instead of 36° and 27° respectively and PQ = 200 instead of 250). Essentially, the working out is the same as long as you use the respective values from your question:
I hope this helps!
Part b:
First we need to find an expression for the PB and QB that involve h
=\frac{PB}{h}\\\\PB=h\cot(36\degree))
Similarly we can deduce that
)
Now using the Cosine Rule
(QB)\cos(74\degree))
Sub in our values for PB and QB
^2+h^2\cot(27\degree)^2-2(h\cot(36\degree))(h\cot(27\degree))\cos(74\degree))
Simplify and factor out
^2+\cot(27\degree)^2-2\cot(36\degree)\cot(27\degree)\cos(74\degree)))
Rearrange to make the subject
^2+\cot(27\degree)^2-2\cot(36\degree)\cot(27\degree)\cos(74\degree)})
For part c just square root both sides
First we need to find an expression for the PB and QB that involve h
Similarly we can deduce that
Now using the Cosine Rule
Sub in our values for PB and QB
Simplify and factor out
Rearrange to make
For part c just square root both sides
Last edited:
One thing that can help is to make it clear in your mind (and thus on the page) which triangle you are using. So, for example:
^2 + \left(h\cot{27^\circ}\right)^2 - 2 \times h\cot{36^\circ} \times h\cot{27^\circ} \times \cos{74^\circ} \\ 250^2 &= h^2\bigg(\cot^2{36^\circ} + \cot^2{27^\circ} - 2\cot{36^\circ}\cot{27^\circ}\cos{74^\circ}\bigg) \\ h^2 &= \cfrac{250^2}{\cot^2{36^\circ} + \cot^2{27^\circ} - 2\cot{36^\circ}\cot{27^\circ}\cos{74^\circ}} \\ \text{Hence,}\ h &= \cfrac{250}{\sqrt{\cot^2{36^\circ} + \cot^2{27^\circ} - 2\cot{36^\circ}\cot{27^\circ}\cos{74^\circ}}} \qquad \text{as $h > 0$} \end{align*})
If necessary, you can follow @may22's suggestion of drawing each of the relevant triangles next to the part of the working dealing with this triangle.
The source of most problems with 3D trig is not the trig, it is extracting the correct triangle in 2D to use for each part.
The source of most problems with 3D trig is not the trig, it is extracting the correct triangle in 2D to use for each part.
Last edited: