Question

paula seeks the width of orion’s belt, a pattern of stars in the orion constellation. the stars alnitak and mintaka are the endpoints of orion’s belt. it is 736 light years (l.y.) from paula’s planet to alnitak and 915 l.y. from her planet to mintaka. the stars are 3° apart in paula’s sky, as shown. what is the width of orion’s belt? that is, what is the distance between alnitak and mintaka? do not round during your calculations. round your final answer to the nearest light year.

Explanation:

Step1: Identify the Law to Use

We have a triangle with two sides and the included angle, so we use the Law of Cosines. The Law of Cosines states that for a triangle with sides \(a\), \(b\), and \(c\), and the included angle \(C\) between sides \(a\) and \(b\), \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\). Here, let \(a = 736\), \(b = 915\), and \(C=3^{\circ}\), and we need to find \(c\) (the distance between Alnitak and Mintaka).

Step2: Apply the Law of Cosines

First, calculate \(a^{2}\), \(b^{2}\), and \(2ab\cos(C)\):

• \(a^{2}=736^{2}=736\times736 = 541696\)
• \(b^{2}=915^{2}=915\times915=837225\)
• \(2ab = 2\times736\times915=2\times673440 = 1346880\)
• \(\cos(3^{\circ})\approx0.9986295348\) (using a calculator for the cosine of 3 degrees)
• Then \(2ab\cos(C)=1346880\times0.9986295348\approx1346880\times0.99863 = 1346880 - 1346880\times(1 - 0.99863)=1346880-1346880\times0.00137 = 1346880 - 1845.2256 = 1345034.7744\)

Now, \(c^{2}=a^{2}+b^{2}-2ab\cos(C)=541696 + 837225-1345034.7744=1378921 - 1345034.7744 = 33886.2256\)

Step3: Find \(c\)

Take the square root of \(c^{2}\) to find \(c\): \(c=\sqrt{33886.2256}\approx184.08\). Rounding to the nearest light year, we get \(c\approx184\).

Answer:

184