p,q and r are points on the same horizontal plane. The bearing of q from p is 150 degree and the bearing of r from q is 060 degree. if pq is 5m and qr is 3m,find the distance of r from q

Answer :

the distance from point \( r \) to point \( p \) is approximately 5.83 meters.

Step-by-step explanation:

To find the distance of \( r \) from \( p \), we need to apply some trigonometry concepts, specifically involving bearings and the Law of Cosines. Here are the steps to solve the problem:

1. **Understand the Bearings:**

- Bearing of \( q \) from \( p \) is 150 degrees.

- Bearing of \( r \) from \( q \) is 60 degrees.

2. **Determine the Angle at \( q \):**

- Bearing angles are measured clockwise from the north direction.

- To find the angle at \( q \), we need to convert these bearings into a common frame of reference and find the angle between them.

- From north, the direction from \( p \) to \( q \) is 150 degrees.

- The direction from \( q \) to \( r \) can be visualized as a turn from the 150-degree line. Since the bearing is 60 degrees, it means 60 degrees clockwise from north (which corresponds to a line oriented at 60 degrees). Therefore, we consider the supplementary angles formed with respect to the 150-degree direction.

- Hence, the interior angle at \( q \) between \( pq \) and \( qr \) is \( 180^\circ - (150^\circ - 60^\circ) = 180^\circ - 90^\circ = 90^\circ \).

3. **Use the Law of Cosines to Find \( pr \):**

- Given:

- \( pq = 5 \) m

- \( qr = 3 \) m

- Angle \( \angle pq r = 90^\circ \)

- Using the Pythagorean theorem (since the angle is 90 degrees, it's a right triangle):

\[

pr = \sqrt{pq^2 + qr^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}

\]

- Therefore, the distance \( pr \) is:

\[

\sqrt{34} \approx 5.83 \text{ meters}

\]

Other Questions