Answer :
To solve for [tex]\( D \)[/tex] given the equation:
[tex]\[ \frac{n}{n_w} = \frac{\sin \left( \frac{D + A}{2} \right)}{\sin \left( \frac{A}{2} \right)} \][/tex]
where [tex]\( n = 1.6 \)[/tex], [tex]\( n_w = \frac{4}{3} \)[/tex], and [tex]\( A = 60^\circ \)[/tex]:
1. Convert angle [tex]\( A \)[/tex] from degrees to radians:
[tex]\[ A = 60^\circ = \frac{60 \pi}{180} = \frac{\pi}{3} \, \text{radians} \][/tex]
2. Calculate [tex]\( \sin \left( \frac{A}{2} \right) \)[/tex]:
[tex]\[ \frac{A}{2} = \frac{\pi}{6}, \quad \sin \left( \frac{\pi}{6} \right) = \sin 30^\circ = \frac{1}{2} \][/tex]
3. Compute the left-hand side of the equation [tex]\( \frac{n}{n_w} \)[/tex]:
[tex]\[ \frac{n}{n_w} = \frac{1.6}{4/3} = 1.6 \times \frac{3}{4} = 1.2 \][/tex]
4. Use the equation to solve for [tex]\( \sin \left( \frac{D + A}{2} \right) \)[/tex]:
[tex]\[ \frac{n}{n_w} = \frac{\sin \left( \frac{D + A}{2} \right)}{\sin \left( \frac{A}{2} \right)} \][/tex]
Substituting the known values:
[tex]\[ 1.2 = \frac{\sin \left( \frac{D + A}{2} \right)}{\frac{1}{2}} \][/tex]
Therefore:
[tex]\[ \sin \left( \frac{D + A}{2} \right) = 1.2 \times \frac{1}{2} = 0.6 \][/tex]
5. Solve for [tex]\( \frac{D + A}{2} \)[/tex]:
[tex]\[ \frac{D + A}{2} = \arcsin(0.6) \][/tex]
To find [tex]\( \arcsin(0.6) \)[/tex], we know that:
[tex]\[ \arcsin(0.6) \approx 0.6435 \, \text{radians} \][/tex]
6. Solve for [tex]\( D \)[/tex]:
[tex]\[ \frac{D + A}{2} = 0.6435 \][/tex]
Therefore:
[tex]\[ D + A = 2 \times 0.6435 = 1.287 \][/tex]
Remember [tex]\( A = \frac{\pi}{3} \approx 1.0472 \)[/tex]:
[tex]\[ D = 1.287 - 1.0472 \approx 0.2398 \, \text{radians} \][/tex]
7. Convert [tex]\( D \)[/tex] from radians to degrees:
[tex]\[ D = 0.2398 \, \text{radians} \times \frac{180}{\pi} \approx 0.2398 \times 57.2958 \approx 13.74^\circ \][/tex]
The nearest given option to [tex]\( 13.74^\circ \)[/tex] is [tex]\( 14^\circ \)[/tex].
Thus, the value of [tex]\( D \)[/tex] is [tex]\( 14^\circ \)[/tex], and the closest option is [tex]\( b \)[/tex].
So, the answer is:
[tex]\[ \boxed{14^\circ} \][/tex]
[tex]\[ \frac{n}{n_w} = \frac{\sin \left( \frac{D + A}{2} \right)}{\sin \left( \frac{A}{2} \right)} \][/tex]
where [tex]\( n = 1.6 \)[/tex], [tex]\( n_w = \frac{4}{3} \)[/tex], and [tex]\( A = 60^\circ \)[/tex]:
1. Convert angle [tex]\( A \)[/tex] from degrees to radians:
[tex]\[ A = 60^\circ = \frac{60 \pi}{180} = \frac{\pi}{3} \, \text{radians} \][/tex]
2. Calculate [tex]\( \sin \left( \frac{A}{2} \right) \)[/tex]:
[tex]\[ \frac{A}{2} = \frac{\pi}{6}, \quad \sin \left( \frac{\pi}{6} \right) = \sin 30^\circ = \frac{1}{2} \][/tex]
3. Compute the left-hand side of the equation [tex]\( \frac{n}{n_w} \)[/tex]:
[tex]\[ \frac{n}{n_w} = \frac{1.6}{4/3} = 1.6 \times \frac{3}{4} = 1.2 \][/tex]
4. Use the equation to solve for [tex]\( \sin \left( \frac{D + A}{2} \right) \)[/tex]:
[tex]\[ \frac{n}{n_w} = \frac{\sin \left( \frac{D + A}{2} \right)}{\sin \left( \frac{A}{2} \right)} \][/tex]
Substituting the known values:
[tex]\[ 1.2 = \frac{\sin \left( \frac{D + A}{2} \right)}{\frac{1}{2}} \][/tex]
Therefore:
[tex]\[ \sin \left( \frac{D + A}{2} \right) = 1.2 \times \frac{1}{2} = 0.6 \][/tex]
5. Solve for [tex]\( \frac{D + A}{2} \)[/tex]:
[tex]\[ \frac{D + A}{2} = \arcsin(0.6) \][/tex]
To find [tex]\( \arcsin(0.6) \)[/tex], we know that:
[tex]\[ \arcsin(0.6) \approx 0.6435 \, \text{radians} \][/tex]
6. Solve for [tex]\( D \)[/tex]:
[tex]\[ \frac{D + A}{2} = 0.6435 \][/tex]
Therefore:
[tex]\[ D + A = 2 \times 0.6435 = 1.287 \][/tex]
Remember [tex]\( A = \frac{\pi}{3} \approx 1.0472 \)[/tex]:
[tex]\[ D = 1.287 - 1.0472 \approx 0.2398 \, \text{radians} \][/tex]
7. Convert [tex]\( D \)[/tex] from radians to degrees:
[tex]\[ D = 0.2398 \, \text{radians} \times \frac{180}{\pi} \approx 0.2398 \times 57.2958 \approx 13.74^\circ \][/tex]
The nearest given option to [tex]\( 13.74^\circ \)[/tex] is [tex]\( 14^\circ \)[/tex].
Thus, the value of [tex]\( D \)[/tex] is [tex]\( 14^\circ \)[/tex], and the closest option is [tex]\( b \)[/tex].
So, the answer is:
[tex]\[ \boxed{14^\circ} \][/tex]