A town Q is on bearing of 210 from P,R is another town on a bearing of 150 from P east of Q.the /RP/=10,find d /RQ/...
I have triangle PQR, with angle P = 60° and RP = 10
not enough information.
triangle PQR is isosceles, since QR runs east-west.
In addition, PQR is equilateral, so RQ=RP=10
To find the distance |RQ| between towns R and Q, we can use the concept of trigonometry and the Sine Rule. Here are the steps to calculate |RQ|:
Step 1: Draw a diagram representing the situation described in the question. Label the points P, Q, and R.
Step 2: Identify and label the given information. We have the bearing from P to Q (210°) and the bearing from P to R (150°). Additionally, we know that |RP| = 10.
Step 3: Notice that triangle PQR is an isosceles triangle since both angles at P are equal (360° - 210° - 150° = 0°). Therefore, |PR| = |PQ|.
Step 4: Use the Sine Rule, which states that for a triangle with sides a, b, and opposite angles A, B, respectively:
a/sin(A) = b/sin(B) = c/sin(C)
In our case, we know that |PR| = |PQ| = c, and the angles opposite these sides are 150° and 210°, respectively. Therefore, we can set up the equation:
|PR|/sin(150°) = |PQ|/sin(210°)
Step 5: Substitute the given values:
10/sin(150°) = |PQ|/sin(210°)
Step 6: Calculate sin(150°) and sin(210°):
sin(150°) = √3/2
sin(210°) = -1/2
Step 7: Substitute the values into the equation:
10/(√3/2) = |PQ|/(-1/2)
Step 8: Simplify the equation:
20/(√3) = -2|PQ|
Step 9: Solve for |PQ| by multiplying both sides by -1/2:
|PQ| = -10√3 / 2
|PQ| = -5√3
However, distance cannot be negative, so we take the absolute value:
|PQ| = 5√3
Therefore, the distance |RQ| between towns R and Q is also 5√3 units.