# Question #03abb

##### 2 Answers

#### Answer:

#### Explanation:

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

Given that:

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

Thus

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using Pythagoras and the given of:

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Thus

We are asked to find:

From the diagram:

#### Answer:

The answer is **(4)**

#### Explanation:

The idea here is that you need to draw the **projection** and the **projection** of the resultant, and use the given **tangent** of the given angle.

Now, you are told that an angle of

#color(red)(alpha) = tan^(-1)(sqrt(3)/2)#

exists between the **resultant**, which we'll call **tangent** of this angle will be

#tan(color(red)(alpha)) = [tan^(1-)(sqrt(3)/2)] = sqrt(3)/2#

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

So,

The **resultant**, **two projections**, one on the

The idea here is that you need to write

Now, focus on the triangle marked with light blue lines. Using the angle

#R_y = Q * sin(color(green)(beta))#

Here *hypotenuse* of the aforementioned triangle.

At the same time, you can say that the **leg** of this triangle located on the

#R_x = P + Q * cos(color(green)(beta))#

Use the given

#{(R_y = 2P * sin(color(green)(beta))), (R_y = P + 2P * cos(color(green)(beta))) :}#

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

#tan(color(red)(alpha)) = sqrt(3)/2#

At the same time,

#tan(color(red)(alpha)) = R_y/R_x#

This means that you have

#sqrt(3)/2 = (2 color(red)(cancel(color(black)(P))) * sin(color(green)(beta)))/(color(red)(cancel(color(black)(P))) + 2color(red)(cancel(color(black)(P))) * cos(color(green)(beta)))#

#sqrt(3)/2 = (2sin(color(green)(beta)))/(1 + 2cos(color(green)(beta)))#

This is equivalent to

#sqrt(3) + 2sqrt(3) * cos(color(green)(beta)) = 4 * sin(color(green)(beta))#

Looking at the options given to you, the only one that matches this equation corresponds to

#sqrt(3) + 2sqrt(3) * cos(60^@) = 4 * sin(60^@)#

#sqrt(3) + color(red)(cancel(color(black)(2)))sqrt(3) * 1/color(red)(cancel(color(black)(2))) = 4 * sqrt(3)/2#

#2sqrt(3) = 2sqrt(3) color(white)(x)color(green)(sqrt())#