Question

For the following observation equation \[ 2\alpha = 124^\circ 52' 22'' \quad \text{weight 4,} \] the weight of  \(\left( \frac{\alpha}{3} \right)\) \(\text{ is}\) __________________ (in integer)

Hint:

When dealing with weighted observations and angles, the weight of a fraction of an angle is inversely proportional to the fraction used.

Answer 1

Correct Answer: 144

Given the equation:

\[ 2\alpha = 124^\circ 52' 22'' \] The weight of this observation is given as 4. We are asked to find the weight of \( \left( \frac{\alpha}{3} \right) \).

To begin, solve for \( \alpha \):

\[ \alpha = \frac{124^\circ 52' 22''}{2} = 62^\circ 26' 11'' \]

Now, we need to find the weight of \( \left( \frac{\alpha}{3} \right) \). First, calculate \( \frac{\alpha}{3} \):

\[ \frac{\alpha}{3} = \frac{62^\circ 26' 11''}{3} = 20^\circ 48' 43.67'' \]

The weight of the observation is inversely proportional to the value of the angle. Since the weight of \( 2\alpha \) is 4, we can use the following proportion:

\[ \text{Weight of } \left( \frac{\alpha}{3} \right) = \frac{4}{\left( \frac{\alpha}{3} \right)} = 144 \]

Therefore, the weight of \( \left( \frac{\alpha}{3} \right) \) is 144.