Question

1. from lighthouses p and q, 16 km apart, a disabled ship s is sighted. if \\(\angle spq = 44^\circ\\) and \\(\angle sqp = 66^\circ\\), find the distance from s to the nearer lighthouse.

Explanation:

Step1: Find the third angle in triangle SPQ

In triangle \( SPQ \), the sum of interior angles is \( 180^\circ \). Given \( \angle SPQ = 44^\circ \) and \( \angle SQP = 66^\circ \), let \( \angle PSQ = x \). Then \( 44^\circ+ 66^\circ + x=180^\circ \), so \( x = 180^\circ-(44^\circ + 66^\circ)=70^\circ \).

Step2: Apply the Law of Sines

The Law of Sines states that \( \frac{SP}{\sin\angle SQP}=\frac{SQ}{\sin\angle SPQ}=\frac{PQ}{\sin\angle PSQ} \). We know \( PQ = 16 \) km, \( \angle SQP = 66^\circ \), \( \angle SPQ = 44^\circ \), \( \angle PSQ = 70^\circ \).
First, find \( SP \) and \( SQ \):

• For \( SQ \): \( \frac{SQ}{\sin44^\circ}=\frac{16}{\sin70^\circ} \), so \( SQ=\frac{16\sin44^\circ}{\sin70^\circ} \). Calculate \( \sin44^\circ\approx0.6947 \), \( \sin70^\circ\approx0.9397 \), then \( SQ\approx\frac{16\times0.6947}{0.9397}\approx11.9 \) km.
• For \( SP \): \( \frac{SP}{\sin66^\circ}=\frac{16}{\sin70^\circ} \), so \( SP=\frac{16\sin66^\circ}{\sin70^\circ} \). Calculate \( \sin66^\circ\approx0.9135 \), then \( SP\approx\frac{16\times0.9135}{0.9397}\approx15.5 \) km.

Step3: Compare the distances

Compare \( SP\approx15.5 \) km and \( SQ\approx11.9 \) km. The smaller one is \( SQ\approx11.9 \) km.

Answer:

The distance from \( S \) to the nearer lighthouse is approximately \( 11.9 \) km (or more precisely, using more accurate sine values, the calculation can be refined, but the approximate value is around 12 km). If we use more precise calculations:
\( \sin44^\circ\approx0.694658 \), \( \sin70^\circ\approx0.939693 \)
\( SQ=\frac{16\times0.694658}{0.939693}=\frac{11.114528}{0.939693}\approx11.83 \) km, and \( SP=\frac{16\times\sin66^\circ}{\sin70^\circ} \), \( \sin66^\circ\approx0.913545 \), so \( SP=\frac{16\times0.913545}{0.939693}=\frac{14.61672}{0.939693}\approx15.55 \) km. So the nearer distance is approximately \( 11.8 \) km (or rounded to 12 km).