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"
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:
$\displaystyle c^2 = a^2 + b^2 - 2ab\cos C$ (where the angle $\displaystyle C$ is opposite to side $\displaystyle c$)
So, we have,
$\displaystyle 18^2 = 20^2 + 33^2 - 2(20)(33)\cos X$
$\displaystyle \Rightarrow324 = 1489 - 1320\cos X$
$\displaystyle \Rightarrow\cos X = \frac{233}{264}$
$\displaystyle \Rightarrow X = \arccos\frac{233}{264}$ (we don't have to worry about other solutions, since $\displaystyle 0 < X < \pi$)
$\displaystyle X\approx0.4895\text{ rad}\approx28.045346^\circ$
Now see if you can convert to deg-min-sec.
Correct!
$\displaystyle 28.045346^\circ = 28^\circ + 0.045346^\circ\left(\frac{60'}{1^\circ}\right)$ (60' per 1°)
$\displaystyle = 28^\circ + \left(0.045346\cdot60\right)' = 28^\circ + 2.72076'$ (canceling units)
$\displaystyle = 28^\circ2' + 0.72076'\left(\frac{60''}{1'}\right)$ (again, multiplying by conversion factor)
$\displaystyle = 28^\circ2' + (0.72076\cdot60)''$ (canceling units)
$\displaystyle \approx28^\circ2'43.25''$ or about $\displaystyle \boxed{28^\circ2'43''}$.