# Question 03abb

Dec 31, 2015

$\angle P A Q = {60}^{o}$

#### Explanation:

The trick is to nearly always draw a diagram and this will help the understanding of what is going on.

Given that: $\textcolor{w h i t e}{. .} {\tan}^{- 1} \left(\frac{\sqrt{3}}{2}\right) = \theta$
and that PC = 2 units marries up very well with triangle PBC. This shows that this diagram is correct.

PC is parallel to AQ so

$\textcolor{b l u e}{\text{Force diagram PAQC reviewed against question and confirmed}}$

$\angle B A Q = \angle B P C = x + \theta \to \textcolor{b l u e}{\text{Checked and confirmed}}$

$\angle \theta = {\tan}^{- 1} \left(\frac{\sqrt{3}}{2}\right) \to \textcolor{b l u e}{\text{ As given and confirmed}}$
Thus $B C = \sqrt{3} \text{ and } B A = 2$

color(magenta)("PA=1 unit from ratio of 2:1") =>PB=1 larr" Updated"#

$\textcolor{b l u e}{\text{Thus BPAC structure confirmed}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using Pythagoras and the given of:
$B C = \sqrt{3} \textcolor{w h i t e}{. .} , \textcolor{w h i t e}{. .} P C = 2$

$B P = \sqrt{{2}^{2} - {\left(\sqrt{3}\right)}^{2}} = 1 \textcolor{b l u e}{\to \text{ BP confirmed}}$

$\textcolor{b l u e}{\Delta \text{ BPC structure confirmed}}$

$\textcolor{red}{\text{Whole diagram is confirmed as correct.}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Thus $\angle {\left(x + \theta\right)}^{o} = \angle B P C \to \textcolor{b l u e}{\text{ Confirmed}}$

We are asked to find: $\angle {\left(x + \theta\right)}^{o}$

From the diagram: $\angle P A Q = {\left(x + \theta\right)}^{o} = {\tan}^{- 1} \left(\frac{\sqrt{3}}{1}\right)$

$\angle P A Q = {60}^{o}$

Jan 1, 2016

The answer is (4) ${60}^{\circ}$

#### Explanation:

The idea here is that you need to draw the $x$-projection and the $y$-projection of the resultant, and use the given $1 : 2$ ratio that exists between $P$ and $Q$ to find the tangent of the given angle.

Now, you are told that an angle of

$\textcolor{red}{\alpha} = {\tan}^{- 1} \left(\frac{\sqrt{3}}{2}\right)$

exists between the resultant, which we'll call $\textcolor{b l u e}{R}$, and $P$. This means that the tangent of this angle will be

$\tan \left(\textcolor{red}{\alpha}\right) = \left[{\tan}^{1 -} \left(\frac{\sqrt{3}}{2}\right)\right] = \frac{\sqrt{3}}{2}$

Keep this in mind. Now, here's a very rough sketch of the two forces

So, $P$ is drawn on the $x$-axis and the angle $\textcolor{red}{\alpha}$ is shown in $\textcolor{red}{\text{red}}$ here.

The resultant, $\textcolor{b l u e}{R}$, will have two projections, one on the $x$-axis, ${R}_{x}$, and the other on the $y$-axis, ${R}_{y}$.

The idea here is that you need to write ${R}_{x}$ and ${R}_{y}$ in terms of $P$, $Q$, and $\textcolor{g r e e n}{\beta}$, the angle that exists between the two forces - shown here in $\textcolor{g r e e n}{\text{green}}$.

Now, focus on the triangle marked with light blue lines. Using the angle $\textcolor{g r e e n}{\beta}$, you can say that

${R}_{y} = Q \cdot \sin \left(\textcolor{g r e e n}{\beta}\right)$

Here $Q$ is the hypotenuse of the aforementioned triangle.

At the same time, you can say that the leg of this triangle located on the $x$-axis will be equal to $Q \cdot \cos \left(\textcolor{g r e e n}{\beta}\right)$, which means that ${R}_{x}$ will be equal to

${R}_{x} = P + Q \cdot \cos \left(\textcolor{g r e e n}{\beta}\right)$

Use the given $1 : 2$ ratio that exists between $Q$ and $P$ to write

$\left\{\begin{matrix}{R}_{y} = 2 P \cdot \sin \left(\textcolor{g r e e n}{\beta}\right) \\ {R}_{y} = P + 2 P \cdot \cos \left(\textcolor{g r e e n}{\beta}\right)\end{matrix}\right.$

Finally, focus on the triangle formed by the two projections of the resultant ans the resultant. You know that

$\tan \left(\textcolor{red}{\alpha}\right) = \frac{\sqrt{3}}{2}$

At the same time,

$\tan \left(\textcolor{red}{\alpha}\right) = {R}_{y} / {R}_{x}$

This means that you have

$\frac{\sqrt{3}}{2} = \frac{2 \textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} \cdot \sin \left(\textcolor{g r e e n}{\beta}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} + 2 \textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} \cdot \cos \left(\textcolor{g r e e n}{\beta}\right)}$

$\frac{\sqrt{3}}{2} = \frac{2 \sin \left(\textcolor{g r e e n}{\beta}\right)}{1 + 2 \cos \left(\textcolor{g r e e n}{\beta}\right)}$

This is equivalent to

$\sqrt{3} + 2 \sqrt{3} \cdot \cos \left(\textcolor{g r e e n}{\beta}\right) = 4 \cdot \sin \left(\textcolor{g r e e n}{\beta}\right)$

Looking at the options given to you, the only one that matches this equation corresponds to $\textcolor{g r e e n}{\beta} = {60}^{\circ}$, since

$\sqrt{3} + 2 \sqrt{3} \cdot \cos \left({60}^{\circ}\right) = 4 \cdot \sin \left({60}^{\circ}\right)$

$\sqrt{3} + \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \sqrt{3} \cdot \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} = 4 \cdot \frac{\sqrt{3}}{2}$

$2 \sqrt{3} = 2 \sqrt{3} \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$