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In RT △ ABC, ∠ C = 90 ° cab = 60 ° and the distance between D and ab is 3.8cm, then how many centimeters is BC The properties of detailed points, especially the right triangle with 30 degree angle
From the title, we can get that de = 3.8cm ? C = 90 °∵ cab = 60 ° B = 30 ° ∵ ad bisection ? cab ? cab = 60 DAB = 30 ° = ∵ B ∵ de ⊥ ab ᚉ e point bisection AB, ab = 2 be ∵ de ⊥ AB, ∵ B = 30 ° be = de ∵ C = 90 ° for 3 times of root number
1. Given that ad is the median line on the hypotenuse BC of the RT triangle ABC, ab = 4, ad = 2.5, then the circumference of the triangle ABC is? 2. If a (A-1) - (a ^ 2-6) = - 5, then a ^ 2 + B ^ 2 / 2-AB =? 3. As shown in the figure is the schematic diagram of fixing the wire pole with the guy wire. Given that CD ⊥ AB, CD = 3 times the root number of 3 meters, and ∠ CAD = ∠ CBD = 60 degrees, then the length of stay wire AC is m
1, if ad is the central line on the hypotenuse BC of the RT triangle ABC, if AB = 2ad = 5 (theorem), then the other side is 3
2. Wrong number
3, ﹤ CAD= ﹤ CBD=60 °, that is, ABC is equilateral triangle
Then ac * (radical 3 / 2) = CD
AC = 6
It is known that ad is the height of the hypotenuse BC of RT △ ABC, BC = α, ∠ B = β, then ad is equal to. (answer: α sin β cos β) process!
Because ad is the height of the hypotenuse BC of RT △ ABC, the triangle ACD is a right triangle, so the angle CAD + angle c = 90 degrees, because the angle B + the angle c = 90 degrees, so the angle CAD = angle B in the right triangle ABC, BC = α, ∠ B = β, so AB = BC * SINB = asin β in the right triangle ACD, so the angle CAD = angle B = β, so ad = AC * cos β = asin β cos β
In RT triangle ABC, CD is the height on the hypotenuse. If ad: BD = 1:2, then the ratio of ACD to BCD is
Easy proof of △ ACD ∽ BCD (projective theorem)
CD×CD=AD×BD
=2
CD=「2
The ratio of perimeter of two triangles = the ratio of sides = 1, "2 = 2:2
As shown in the figure, if the circumference of △ ABC is 32, and ab = AC, ad ⊥ BC is in D, and the circumference of ⊥ ACD is 24, then the length of ad is______ .
∵AB=AC，AD⊥BC，
∴BD=DC．
∵AB+AC+BC=32，
AB + BD + CD + AC = 32,
∴AC+DC=16
∴AC+DC+AD=24
∴AD=8．
Therefore, fill in 8
It is known that in the RT triangle ABC, the angle c is equal to 90 degrees, point D is the midpoint of AB, e is any point on AC, and DF is vertical de intersecting BC with point F Verification: "AE ^ 2 + BF ^ 2 = CE ^ 2 + CF ^ 2“
Connect EF, rotate the triangle ade 180 degrees anticlockwise around point d to obtain △ BDE '(AD coincides with BD) to connect Fe'
Because the angle FDE = 90 ° so the angle ade + angle BDF = 90 ° because the angle ade = angle BDE ', so the angle FDB + angle BDE' = 90 ° EDE 'are collinear. Because de = de', FD ⊥ EE ', so Fe = Fe'
Because AE = be ', AE ^ 2 + BF ^ 2 = be' ^ 2 + BF ^ 2 = e'f ^ 2
Because ∠ C = 90 °, CE ^ 2 + CF ^ 2 = EF ^ 2
So AE ^ 2 + BF ^ 2 = CE ^ 2 + CF ^ 2
As shown in the figure, CD is the height on the hypotenuse ab of RT △ ABC. Fold △ BCD along CD, and point B just falls at the midpoint e of AB, then ∠ A is equal to______ Degree
∵ in RT △ ABC, CE is the center line of the hypotenuse ab,
∴AE=CE，
∴∠A=∠ACE，
∵ △ CED is folded from △ CBD,
∴∠B=∠CED，
∵∠CEB=∠A+∠ACE=2∠A，
∴∠B=2∠A，
∵∠A+∠B=90°，
∴∠A=30°．
So the answer is: 30
It is known that the radius of circle 0 is 2, the length of chord AB is 2 √ 3, point C and point D are any points on inferior arc AB and superior arc ADB respectively It is known that the radius of circle 0 is 2, the length of chord AB is 2 √ 3, point C and point D are any points on inferior arc AB and superior arc ADB respectively. Find the maximum areas of angle ACB and triangle abd
Angle ACB = 180 (12-3) / 12
=135°
The maximum area of angular abd = 2 √ 3 * 3 / 2
=3√3
As shown in the figure, the radius of circle O is 6, the chord AB = 6, the root sign 3, C is a point on the circle, C does not coincide with a and B, calculate the degree of ∠ ACB
1. I don't see the picture. According to my understanding, make C on the inferior arc of ab
connect OA.OB
Through O, make OD ⊥ AB at point D
Sin ∠ DOB = 1 / 2Ab / ob = 3 root 3 / 6 = root 3 / 2
∠DOB=60°
∠AOB=120°
∠ACB=1/2∠AOB=60°
2 C on the inferior arc of ab
∠ACB=1/2（360°-120°）=120°
It is known that the radius of circle O is 1, AB is a chord of circle O, and ab = radical 3 AOB () A. 30 ° b.60 ° c.30 ° or 150 ° d.60 ° or 120 ' Seeking process 120 ° I can understand.. What is the 60 ° situation?
Make a vertical line co to ab
Because AB is a chord of circle o
So AC = BC = 0.5 times root number 3
And the radius of circle O is 1
So the angle AOC is 60 degrees
Then ∠ AOB = 120 degrees
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