Answer:
(i) 20 meters
(ii) 8.0 meters (nearest tenth)
Step-by-step explanation:
A drawbridge is 60 meters long when stretched across a river. As the bridge opens, it does so in the middle, and the two sections of the bridge, each 30 meters long, can be rotated upward through an angle of 30°.
Part (i)
When the two bridge sections are rotated so the bridge is fully open, they form two 30-60-90 right triangles. The hypotenuse is the rotated section (30 meters), and the height of the triangle is the shortest leg.
In a 30-60-90 triangle, the leg opposite the smallest angle is half the length of the hypotenuse. Therefore, the vertical distance between the end of the opened bridge section and the horizontal line of the closed bridge is 15 meters.
As the water level is 5 meters below the closed bridge, the total height between the end of a section and the water level when the bridge is fully open is:
[tex]\begin{aligned}h&=15\; \textsf{meters} + 5\; \textsf{meters}\\h&= 20\; \textsf{meters}\end{ailgned}[/tex]
Part (ii)
To calculate the distance between the ends of the two sections when the bridge is fully opened, the 30-60-90 right triangle property can be used again.
The base of the right triangle is opposite the 60° angle, making it √3 times the length of the leg opposite the 30° angle. Therefore, the base of the right triangle is 15√3 meters.
Given that the width of the bridge when closed is 60 meters, the distance between the ends of the two sections when the bridge is fully open is the length of the closed bridge minus two bases of the right triangle:
[tex]\begin{aligned}\textsf{Distance}&=60 - 2(15\sqrt{3})\\\textsf{Distance}&= 60 - 30\sqrt{3}\\\textsf{Distance}&= 8.03847577... \\\textsf{Distance}&=8.0 \; \sf meters\; (nearest\; tenth)\end{aligned}[/tex]
