I need to calculate the unknown quantity x and express answer in degrees, minutes & seconds. Pls advise if my calculations are correct and if not, could please assist me - thanks again.

tan^-1 (20/18)
x = 48.0128

x = 48^o, 0' & 46"

2. Originally Posted by Risrocks
I need to calculate the unknown quantity x and express answer in degrees, minutes & seconds. Pls advise if my calculations are correct and if not, could please assist me - thanks again.

tan^-1 (20/18)
x = 48.0128

x = 48^o, 0' & 46"
No. Be very careful here. You can only use the trigonometric ratios when dealing with right triangles, and this is not a right triangle.

Here, we use the law of cosines:

$c^2 = a^2 + b^2 - 2ab\cos C$ (where the angle $C$ is opposite to side $c$)

So, we have,

$18^2 = 20^2 + 33^2 - 2(20)(33)\cos X$

$\Rightarrow324 = 1489 - 1320\cos X$

$\Rightarrow\cos X = \frac{233}{264}$

$\Rightarrow X = \arccos\frac{233}{264}$ (we don't have to worry about other solutions, since $0 < X < \pi$)

$X\approx0.4895\text{ rad}\approx28.045346^\circ$

Now see if you can convert to deg-min-sec.

3. Thanks - I should have dealt with it as an obtuse triangle.

Converted = 28^o 2' 43"

Thanks for taking the time to assist me.

4. Originally Posted by Risrocks
Thanks - I should have dealt with it as an obtuse triangle.

Converted = 28^o 2' 43"

Thanks for taking the time to assist me.
Correct!

$28.045346^\circ = 28^\circ + 0.045346^\circ\left(\frac{60'}{1^\circ}\right)$ (60' per 1°)

$= 28^\circ + \left(0.045346\cdot60\right)' = 28^\circ + 2.72076'$ (canceling units)

$= 28^\circ2' + 0.72076'\left(\frac{60''}{1'}\right)$ (again, multiplying by conversion factor)

$= 28^\circ2' + (0.72076\cdot60)''$ (canceling units)

$\approx28^\circ2'43.25''$ or about $\boxed{28^\circ2'43''}$.