Question

Question #03abb

Answer 1

`/_ PAQ =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: `color(white)(..)tan^(-1)(sqrt(3)/2) = 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

`color(blue)("Force diagram PAQC reviewed against question and confirmed")`

`/_BAQ = /_BPC = x+theta->color(blue)("Checked and confirmed")`

`/_ theta =tan^(-1)(sqrt(3)/2)-> color(blue)(" As given and confirmed")`
Thus `BC=sqrt(3)" and "BA=2`

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

`color(blue)("Thus BPAC structure confirmed")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using Pythagoras and the given of:
`BC=sqrt(3)color(white)(..) ,color(white)(..) PC = 2`

`BP = sqrt( 2^2 - (sqrt(3))^2) = 1 color(blue)(->" BP confirmed")`

`color(blue)(Delta" BPC structure confirmed")`

`color(red)("Whole diagram is confirmed as correct.")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Thus `/_ (x+theta)^o = /_BPC ->color(blue)(" Confirmed")`

We are asked to find: `/_ (x+theta)^o`

From the diagram: `/_ PAQ = (x+theta)^o = tan^(-1)(sqrt(3)/1)`

`/_ PAQ =60^o`

Answer 2

The answer is (4) `60^@`

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

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

exists between the resultant, which we'll call `color(blue)(R)`, and `P`. This means that the 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, `P` is drawn on the `x`-axis and the angle `color(red)(alpha)` is shown in `color(red)("red")` here.

The resultant, `color(blue)(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 `color(green)(beta)`, the angle that exists between the two forces - shown here in `color(green)("green")`.

Now, focus on the triangle marked with light blue lines. Using the angle `color(green)(beta)`, you can say that

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

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 * cos(color(green)(beta))`, which means that `R_x` will be equal to

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

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

`{(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 `color(green)(beta) = 60^@`, since

`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())`