Ask a math question about solving a right triangle to solve the problem in detail ! Urgent

Solution: It is known that f is the point where AD edge pairs are superimposed on BC.

then afe=90°

1 + 2 = 90° (this doesn't need to be explained) and c = 90° 2 + 3 = 90°, the same goes for 1 + 4 = 90°

abf ∽ fce

It is also known that tan efc = 3 4

tan∠efc=ce/fc=3/4

Let ce=3x and fc=4x

According to the Pythagorean theorem, fe=5x

de=ef ∴dc=de+ec=8x

Proportional to the corresponding sides of similar triangles.

CE BF = FC AB

i.e. 3x bf =4x 8x

Get bf = 6x

bc =bf+fc=10x

According to the Pythagorean theorem.

ae² =ad² +de²

i.e. (5 times the root number 5 ) = (10x) 5x) solution gives x=1 ad=bc=10 ab=dc=8c = (10+8) 2=36

If CE is 3x cm long, then CF = 4x cm.

So ef = 5x cm.

So ab=cd=8x cm.

Because the angle AFB + angle EFC = 90 degrees and the angle EFC + angle FEC = 90 degrees.

So the triangle ABF is similar to the triangle FCE

And because ab:cf=8x:4x=2:1

So bf = 2ce = 6x

So af=10x and ef=5x

So ae = 5 root number 5x

And because ae = 5 times the root number 5 cm.

So x=1 so cd=8 cm, bc=10 cm.

So the quadrilateral circumference is 36 cm (sort it out yourself).

Suppose the point in the corner of the wall is p, and the length of a1p is h

then ap = h+1

In the triangle apb, tan = 4 3, so sin = 4 5, so ab = (h+1) *5 4

Look at the triangle a1pb1 again, sin = h a1b1 = 3 5, and a1b1 = ab

So ab=h*5 3, combine the two ab columns to get :

5/3 * h = 5/4 * h+1)

The solution is h=3, ab=5

So the ladder length is 5m

It's actually quite simple.

2*(2 tan30 degrees)*30 + 2*2*30

Do the math yourself.

First, calculate the length of the stair ramp: l = 2 sin30° = 4 m.

Then calculate the slope area of the staircase: s=2*4=8 square meters.

8*30=240 yuan.

A minimum of $240 is required.

This process should be very clear, the answer is as follows: dam bottom length: 6 + 10 2 = 11 meters cross-sectional area: (6 + 11) * 10 2 = 85 square meters after heightened bottom length: 11 + 2 2 = 12 meters.

After heightening, the cross-sectional area: (12 + 6) * 12 2 = 108 square meters area increase: 108-85 = 23 square meters.

Reinforcement of 50 meters of soil is required: 23 * 50 = 1150 cubic meters.

Dam bottom length: 6 + 10 2 = 11 meters.

Cross-sectional area: (6+11)*10 2=85 square meters, heightened bottom length: 11+2 2=12 meters.

After heightening, the cross-sectional area: (12 + 6) * 12 2 = 108 square meters area increase: 108-85 = 23 square meters.

Reinforcement of 50 meters of soil is required: 23 * 50 = 1150 cubic meters.

Hehe, it's a little stupid, and I still haven't made it after half an hour.

1) Make the midpoint of AB (there are many ways to use the regular triangle method), take the midpoint as the perpendicular foot, make a perpendicular line in the direction of AC, and cross AC at the point P, which is what is sought.

2) A+ abc=90°, abp= a, so 2 A+ PBC=90°, so BPC=90°- PBC=2 A=51°10

Since acb=45°, we can first set ab=bc=x and adb=30°, so ab bc= 3 3x (x+100)= 3 3 solution x=50( 3+1) is desirable.