The triangle ABC is a right triangle, the angle B = 90 degrees, the angle c = 30 degrees, BC = 5 times the root sign 3 E and f start from AB and Ca respectively and move along the direction of AB and ca. point E is 1cm per second, point F is 2cm per second, and FD is made through point F perpendicular to BC to D. It is proved that the quadrilateral afdh is a flat quadrilateral. When the movement time is how many, the triangle EFD right triangle is right triangle You can't use sine cosine. I haven't learned it yet

It is proved that if the motion is t second, so CF = 2tcm, AE = TCM ∵ FD ⊥ BC ᚉ ab ∵ FD ? C = 30  FD = fcsin  C = 2tsin30 ° = t = AE  the quadrilateral afdh is a parallelogram because de ^ 2 = (be ^ 2 + BD ^ 2) = 4 (T ^ 2-10t + 25), EF ^ 2 = 7T ^ 2-50t + 100, FD ^ 2 = T ^ 2, when de ^ 2 = EF ^ 2 + FD ^ 2, t =... "

In the right triangle ABC, the angle c is equal to 90 degrees and the angle B is equal to 60 degrees. What is the ratio of the lengths of BC, AC and ab In right triangle ABC, if angle c equals 90 degrees and angle B equals 60 degrees, what is the ratio of length BC, AC and ab If BC equals root 3, then what is AC and what is ab

Draw a triangle
AC / AB = radical 3 / 2
AC = 3 AB = 2 roots 3

ABC right triangle. AB is 80 mm long

The angle B = 90 ° indicates a right triangle
So BC = AB / Tanc = 80 / tan70 ° = 29mm
BC length is 29mm
Tan 48 ° = 1.11 (by calculator only)

Given the right triangle ABC, the angle B is 90 ° and ab is 7 cm, 2BC = AC, find the length of BC?

Junior high school method: set BC as X and AC as 2x
∴x²+7²=（2x）²
The root of x = 7 / 3 is 3cm
High school method: BAC = 30 ° from the meaning of the title
/ / BC = Tan 30 ° times AB = 7 / 3 of the root 3 cm

The area of the two shadow parts in the figure is equal. The triangle ABC is a right triangle and BC is a diameter of 40 cm. Find the length of ab

Where is the picture?

As shown in the figure: in RT △ ABC, ab = AC, ∠ BAC = 90 ° and O is the midpoint of BC (1) Write the relationship between the distance between point O and the three vertices a, B and C of △ ABC; (2) If the point m and N move on the segments AB and AC respectively, and an = BM is maintained during the movement, please judge the shape of △ omn and prove your conclusion

(1) ∵ in RT △ ABC, ∵ BAC = 90 ° and O is the midpoint of BC,
∴OA=1
2BC=OB=OC，
OA = ob = OC;
(2) The △ omn is an isosceles right triangle
Connect Ao
∵AC=AB，OC=OB
∴OA=OB，∠NAO=∠B=45°，
In △ AON and △ BOM
AN＝BM
∠NAO＝∠B
OA＝OB
∴△AON≌△BOM（SAS）
∴ON=OM，∠NOA=∠MOB
∴∠NOA+∠AOM=∠MOB+∠AOM
∴∠NOM=∠AOB=90°，
The △ omn is an isosceles right triangle

The oblique side ab of the RT triangle ABC is in plane a. the angles formed by AC, BC and a are 30 ° and 45 ° respectively. CD is the high line on ab. find the angle formed by Cd and plane a Contents of PEP B Edition Anybody?

The specific solution can be seen in the graphic reference:
60°
Make CE ⊥ plane α, connect AE, EB, ed
CE ⊥ plane α - - > CE ⊥ AE, CE ⊥ EB, CE ⊥ ed
The angle CAE is the angle formed by AC and α
The angle EBE is the angle formed by BC and α
Angle CAE = 30 degrees
Angle EBE = 45 degrees
Let CE = X
AC=2x,CB=sqrt(2)x
AC⊥BC
AB=sqrt(6)x
CD = (2sqrt(2)/sqrt(6))x
DE⊥AB
The angle CDE is the angle formed by Cd and plane α

It is known that the hypotenuse ab of the RT triangle ABC is in the plane Z, and the angles formed by AC, BC and plane Z are 30 and 45 degrees, respectively

It should be 60 ° to solve the problem. The train of thought is very clear to make the CD ⊥ plane Z, the foot of the plumb is D, connect AD and BD, then ﹣ CAD = 30 °, CBD = 45 ° and set CD = a, we can calculate the length of AC, ad, BD, BC, AB as de ⊥ AB, connect CE, it is very simple to deduce ab ⊥ CE, then ﹣ CED is the required angle setting ﹣ PAB = α according to the formula de ⊥ = ad #

The hypotenuse BC of RT triangle ABC is in plane M. the angles formed by two right angles and plane m are 45 ° and 30 ° respectively Find the angle between the height ad of the hypotenuse and the plane M

Make AE ⊥ plane m, connect CE, EB, ed
AE ⊥ plane m --- > AE ⊥ be, AE ⊥ EC, AE ⊥ ed
The angle between AB and m and ACE is the angle between AC and m
∴∠ABE=30º, ∠ACE=45º
Let AE = x, then AB = 2x, CA = √ 2x
AB⊥AC, ∴BC=√6x AD= (2/√3)x
De ⊥ BC,  ade is the angle formed by AD and plane M
∴sin∠ADE=AE/AD= x/(2/√3)x=√3/2===>∠ADE=60º
The angle formed by the height ad of the beveled edge and the plane M = 60 °

In the RT triangle ABC, if angle B = 90 degrees, angle c = 30 degrees, D is the midpoint of BC, AC = 2, de ⊥ plane ABC, and de = 1, then the distance from point e to hypotenuse AB is? It's bevel AC

If the intersection of point E and AC is f, and point F should be a straight line passing through point d to make the hypotenuse AC. then EF = 1 / 2ec = 1 / 4bc (the right angle side corresponding to 30 ° is half of the length of the hypotenuse) = (√ 3 / 4). The triangle EDF is a right triangle, CF is a hypotenuse. According to the Pythagorean theorem, = (√ 19 / 4), then the distance from point e to the hypotenuse AC is) = √ 19 / 4