Please help me. P and Q are two gun placements 1250m apart. T is a target. The angles TPQ and TQP are 73 degree 24' and 68 degree 5' respectively. Find the distance P to the target T.
you know that angle T is 8° 31'
PT/sinQ = 1250/sinT
Hmmm. Bit of a typo there. T is 38°31'
To solve this problem, we can use trigonometry, specifically the Law of Sines. The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant.
Let's label the distance from P to T as x. Now, we can draw a triangle with sides PT, QT, and PTQ.
We are given that PTQ = 68° 5' and TPQ = 73° 24'.
To use the Law of Sines, we need to find the sine of the angles.
sin(68° 5') = sin(68 + 5/60) = sin(68.083°) ≈ 0.9329
sin(73° 24') = sin(73 + 24/60) = sin(73.4°) ≈ 0.9536
Now, we can apply the Law of Sines:
PT/sin(TPQ) = QT/sin(TQP)
PT/sin(73.083°) = 1250m/sin(68.4°)
Let's solve this equation for PT:
PT = sin(73.083°) * (1250m/sin(68.4°))
PT ≈ 0.9329 * (1250m/0.9536)
PT ≈ 1221.73m
Therefore, the distance from P to the target T is approximately 1221.73 meters.