The original height can be expressed as 4Xsin40.
Let the adjusted length be x, which is xXsin35.
4Xsin40=xXsin35
X = 4.48m m.
That is, it has grown by 0.48 meters.
It turns out that the length of stairs occupying the ground is 4Xcos40.
The adjusted floor length is 4.48Xcos35.
That is 4.48 xcos35-4x cos40 = 0.61m.
Classroom exercises 1:
BC = BDXTAN40 = 5xTAN40 = 4.20m。
Be = BC+CE = 4.20+2 = 6.20m。
pythagorean theorem
DE= root number (6.20 square +4 square) = root number 54.44 = 7.38m
The cable length is 7.38m.
Classroom exercise 2:
Angle ADC= 135 can be used to deduce the angle DCB=45.
Do high from D to BC, cross BC at E, and cross BC at F from A to BC.
Af = ce = de = 8x sin45 = 5.66m。
BF = 30-6-5.66 = 18.34m。
tanABC = AF/BF = 5.66/ 18.34 = 0.3 1
Angle ABC= 17.22 degrees.
Trapezoidal area = (30+6) x 5.66/2 =101.88m2.
Dam volume = trapezoidal area x length =101.88x100 =10/88.00 m3.
That is, earthwork 10 188.00 cubic meters is required.