A 5mi straight segment of a road climbs a 4000ft hill Determ

A 5-mi straight segment of a road climbs a 4000-ft hill. Determine the angle that the road makes with the horizontal. A box has dimensions as shown in FIGURE 5.2.10. Find the length of the diagonal between the corners P and Q. What is the angle theta formed between the diagonal and the bottom edge of the box? Observers in two towns A and B on either side of a 12,000-ft mountain measure the angles of elevation between the ground and the top of the mountain. See FIGURE 5.2.11. Assuming that the towns and the mountaintop lie in the same vertical plane, find the horizontal distance between them. A drawbridge* measures 7.5 m from shore to shore, and when completely open it makes an angle of 43 degree with the horizontal. See FIGURE 52.12(a). When the bridge is closed, the angle of depression from the shore to a point on the surface of the water below the opposite end is 27 degree. See Figure 5.2.12(b). When the bridge is fully open, what is the distance d between the highest point of the bridge and the water below?

Solution

The diagonal marked there is a SPACE DIAGONAL
which passes through the bottom right endpoint of the back surface
to the top left endpoint of the front surface

The space diag = sqrt(a^2 + b^2 + c^2)

d = sqrt(4^2 + 3^2+ 3^2)

d = sqrt(34) ----> ANSWER

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For theta, the opposite side = sqrt(3^2 + 3^2), which is the face diagonal

opp = sqrt18

opp = 3sqrt(2)

and hyp = sqrt34

So, sin(theta) = 3sqrt(2) / sqrt(34)

sin(theta) = 0.7276068751089989

theta = 46.686 degrees -----> ANSWER