1. ## rectangular section PQRS

hi to all i would like some help on the that math project

2. Note that angle QAB has measure 90 - x.

Also, cos(QAB) = QB/AB = QB/18, so QB = 18(cos(90-x)).

Similarly, cos(CBR) = BR/27, so BR = 27(cos(x)).

Note that QB + BR = 30 (the bottom part of your diagram).

So, 18(cos(90-x)) + 27(cos(x)) = 30.

Use the fact that the graph of cosine is the same as the graph of sine (just shifted) to convert cos(90-x) to sin(x).

Finally, divide everything by 3.

3. Hello,

Note that angle QAB has measure 90 - x.
I'd rather say that it's angle QBA whose measure is 90-x.

Angle QAB has measure x

a) Prove that: .$\displaystyle 9\cos x + 6\sin x \:=\:10$
We have: .$\displaystyle AB = DC = 18$
Note that: .$\displaystyle \angle BAQ = x$

In right triangle $\displaystyle AQB\!:\;\;\sin x \,=\,\frac{QB}{18}\quad\Rightarrow\quad QB \,=\,18\sin x$

We have .$\displaystyle BC = AD = 27$
In right triangle $\displaystyle CRB\!:\;\;\cos x \:=\:\frac{BR}{27}\quad\Rightarrow\quad BR \,=\,27\cos x$

Since $\displaystyle QR \:=\:BR + QB \:=\:30$

. . we have: .$\displaystyle 27\cos x + 18\sin x \:=\:30 \quad\Rightarrow\quad\boxed{ 9\cos x + 6\sin x \:=\:10}$

b) Express $\displaystyle 9\cos x + 6\sin x$ in the form: $\displaystyle R\cos(x - a)$
. . where $\displaystyle R > 0\text{ and }0^o < a < 90^o,$
. . giving the values of $\displaystyle R\text{ and }a$ to 2 decimal places.
This takes a bit of work . . .
If you've never seen this done, no wonder you are baffled.

We have: .$\displaystyle 9\cos x + 6\sin x$

. . Find: .$\displaystyle \sqrt{9^2+6^2} \;=\;\sqrt{117} \:=\:3\sqrt{13}$

Multiply and divide by $\displaystyle 3\sqrt{13}\!:\;\;3\sqrt{13}\,\left[\frac{9}{3\sqrt{13}}\cos x + \frac{6}{3\sqrt{13}}\sin x\right]$

. . and we have: .$\displaystyle 3\sqrt{13}\,\left(\frac{3}{\sqrt{13}}\cos x + \frac{2}{\sqrt{13}}\sin x\right)$

Let $\displaystyle \cos a \,=\,\frac{3}{\sqrt{13}}\,\text{ and }\,\sin a \,=\,\frac{2}{\sqrt{13}}$

. . and we have: .$\displaystyle 3\sqrt{13}\,\left(\cos a\cos x + \sin a\sin x\right) \;=\;\boxed{3\sqrt{13}\,\cos(x-a)}$

Hence: .$\displaystyle R\:=\:3\sqrt{13} \quad\Rightarrow\quad\boxed{R \:\approx\:10.82}$

And: .$\displaystyle a \,=\,\cos^{-1}\left(\frac{3}{\sqrt{13}}\right) \quad\Rightarrow\quad\boxed{a \:\approx\:33.69^o}$