P and Q are two observation posts on the same horizontal ground as the foot R of a vertical pole RT. P is 2.5km south of R. Q is east of R and is on a bearing of 065° from P. The angle of elevation of T from P is 40°. The distance RT is 2.19km.The distance RQ is 2.079km and the distance PQ is 0.977km.Calculate,to three significant figures angle of elevation of T from Q
sorry - I cannot believe the pole RT is 2.19 km high.
Something is wrong with this problem.
and your diagram is also wrong, since P is south of R
To calculate the angle of elevation of T from Q, we need to use the concept of triangle trigonometry.
Let's first draw a diagram to visualize the given information:
P(2.5 km)
/ \
/ \
R -------------- Q(2.079 km)
\ /
\ /
T(2.19 km)
From the given information, we can deduce that angle TPQ is a right angle (since T is the top of the pole and is vertically above point R).
Therefore, angle PTQ is 90 degrees.
Now, we can use the law of cosines to find angle PTQ:
cos(PTQ) = (PQ^2 + TP^2 - TQ^2) / (2 * PQ * TP)
Substituting the given values:
cos(PTQ) = (0.977^2 + 2.19^2 - 2.079^2) / (2 * 0.977 * 2.19)
cos(PTQ) = (0.9548502401 + 4.7961 - 4.3240241) / (4.28742)
cos(PTQ) = (1.42692524) / (4.28742)
cos(PTQ) = 0.332952097941
Now, we can find angle PTQ by taking the inverse cosine (cos^-1) of the value we just calculated:
PTQ = cos^-1(0.332952097941)
PTQ ≈ 70.909
Therefore, the angle of elevation of T from Q is approximately 70.909 degrees.
Apologies for the confusion. I made an error in interpreting the problem. Let's solve it correctly.
R(2.5 km)
/ \
/ \
P -------------- T(2.19 km)
\ /
\ /
\ /
\ /
Q(2.079 km)
Given that P is 2.5 km south of R, we can assume it is directly below R on the same vertical line. Therefore, the angle of elevation of T from P is 40 degrees.
We need to find the angle of elevation of T from Q. To do this, we will use the concept of triangle trigonometry, specifically the tangent function:
tan(angle) = opposite / adjacent
In this case, the opposite side is TP (height of T above P) and the adjacent side is PQ.
We know that PQ is 0.977 km and TP is the height of the pole above P, which we need to determine. We can calculate TP using the tangent of the angle:
tan(40°) = TP / 0.977
TP = 0.977 * tan(40°)
TP ≈ 0.977 * 0.8390996311
TP ≈ 0.8188 km
Now, we have PQ = 0.977 km and TP = 0.8188 km. We can use the Pythagorean theorem to find TQ:
TQ^2 = PQ^2 + TP^2
TQ^2 = 0.977^2 + 0.8188^2
TQ^2 = 0.954649 + 0.67021144
TQ^2 ≈ 1.6249
TQ ≈ √(1.6249)
TQ ≈ 1.2767 km
Now, we have TQ = 1.2767 km and PQ = 0.977 km. We can use the tangent function again to find the angle of elevation of T from Q:
tan(angle) = opposite / adjacent
tan(angle) = TQ / PQ
angle = tan^-1(TQ / PQ)
angle ≈ tan^-1(1.2767 / 0.977)
angle ≈ 53.95°
Therefore, the angle of elevation of T from Q is approximately 53.95 degrees.