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a traveller moves from a town p on a bearing of 055° to a town q 200km …

Question

a traveller moves from a town p on a bearing of 055° to a town q 200km away. he then moves from q on a bearing of 155° to a town r, 400km from q. find the distance between p and r.

Explanation:

Step1: Calculate the angle at \( Q \)

The bearing from \( P \) to \( Q \) is \( 055^{\circ} \), and the bearing from \( Q \) to \( R \) is \( 155^{\circ} \).
The angle \( \angle PQR=155^{\circ}- 55^{\circ}=100^{\circ}\)

Step2: Apply the Law of Cosines

The Law of Cosines states that for a triangle with sides \( a,b,c \) and the angle \( C \) opposite side \( c \), \( c^{2}=a^{2}+b^{2}-2ab\cos C\)
Here, \( a = 200\), \( b = 400\), and \( C = 100^{\circ}\), \( \cos(100^{\circ})\approx - 0.1736\)

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Step3: Find the value of \( PR \)

Take the square - root of \( PR^{2}\)
\( PR=\sqrt{227776}\approx477.3\)

Answer:

The distance between \( P \) and \( R \) is approximately \( 477.3\) km.