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In the triangle ABC, the angle c = 90 degrees, D is a point on the edge AC, and ad = 1, angle abd = 45 degrees. Let the angle CBD = a, BC = X. when Tana = 3 / 5, find x =? When x = 0.16, find Tana =?
(1) When Tana = 3 / 5, but in RT triangle BCD, Tana = CD / BC, CD = 3k, BC = 5K (k > 0)
In RT triangle ABC: Tan (a+45 °) =AC/BC
→ 1+tana 1+3k
__________ = __________
1-tana 5k
→ 1+3/5 1+3k
_________ = ______________
1-3/5 5k
→k=1/17
∴x=5k=5×(1/17)=5/17
(2) In RT △ BCD, Tana = CD / BC, CD = bctana = 0.16tana
In RT △ ABC: Tan (a + 45 °) = AC / BC
→ 1+tana 1+0.16tana
———— = ———————
1-tana 0.16
→0.16tan²a+tana-0.84=0
→4tan²a+25tana-21=0
→（4tana-3）（tana+7）=0
Tana = 3 / 4 or Tana = - 7
Because a can only be an acute angle, Tana > 0
∴tana=3/4
As shown in the figure, in △ ABC, ∠ ABC = 90 °, BD ⊥ AC in D, ∠ CBD = α, ab = 3, BC = 4, find the values of sin α, cos α, Tan α
From the question: ∠ a = ∠ CBD = α
Therefore, Tan α = Tana = A / C = AB / BC = 3 / 4
Therefore, sin α = 3 / 5, cos α = 4 / 5
If BD = 8, sin angle CBD = 0.75, find the length of AE
DF ⊥ BC is made by D at point F
Sin angle CBD = DF / BD = 0.75, so DF = bdsin angle CBD = 8 * 0.75 = 6
However, AE ⊥ BC, DF / / AE, △ CDF ∽ CAE
Therefore: DF / AE = CD / AC, we can see: AE = AC * DF / CD = 6 * 3 / 2 = 9
As shown in the figure, D is a point on the edge AC of △ ABC, CD = 2ad, AE ⊥ BC in E, if BD = 8, sin ≁ CBD = 3 4. Find the length of AE
If DF ⊥ BC is used as DF ⊥ BC in F, then DF = BD · sin ∠ CBD = 8 × 3
4=6，
From AE ⊥ BC, DF ⊥ BC, DF ⊥ AE,
So DF
AE=CD
AC=2
Therefore, AE = 3
2DF=9．
As shown in the figure, in △ ABC, ∠ BAC = 90 °, D is the midpoint of AC, AE ⊥ BD, e is the perpendicular foot, and it is proved that ∠ CBD = ∠ ECD
It is proved that ∵ in △ ABC, ∵ BAC = 90 ° AE ⊥ BD,
∴∠AED=∠BAD=90°，
∵∠ADE=∠BDA，
∴△ADE∽△BDA，
∴AD：BD=DE：AD，
∵ D is the midpoint of AC,
∴AD=CD，
∴CD：BD=DE：CD，
∵∠CDE=∠BDC，
∴△CDE∽△BDC，
∴∠CBD=∠ECD．
In triangle ABC, angle ACB = 90 °, AE = AC, BD = BC, calculate the degree of triangle ECD E. D on ab
set up
In the triangle ABC, if the vertical bisector De of AB intersects AC at D and ab at e, given AC = 5 and BC = 4, then the circumference of the triangle BDC is
Because BD = ad, the perimeter is 9
As shown in the figure, in △ ABC, De is the median perpendicular of AC, which intersects AC and ab at points D and e respectively. Given AB + BC = 6cm, calculate the circumference of △ BCE
∵ De is the perpendicular of AC,
∴AE=CE，
The circumference of △ BCE =CE+BE+BC=AE+BE+BC=AB+BC,
∵AB+BC=6cm，
The circumference of △ BCE = 6cm
In the triangle ABC, the angle ACB is equal to 90 degrees, the vertical bisector De of BC intersects BC on D, foot AB on e, f on De, and AF = CE It is proved that ACEF is a parallelogram
∵ the vertical bisector BC of De, CED = ∠ bed
∵FD‖AC,∴∠CED=∠ACE,∠CAE=∠BED,
∴∠ACE=∠CAE,∴CE=AE
And AF = CE, AE = AF, AFE = ∠ AEF = ∠ bed = ∠ CED
∵∠CEF+∠CED=180º,
∴∠CEF+∠AFE=180º,∴CE‖AF
And ∵ CE = AF
The ACEF is a parallelogram
As shown in the figure, in △ ABC, ∠ ACB = 90 °, the vertical bisector of BC intersects BC at D, AB at point E, f on De, and AF = CE （1） Verification: quadrilateral ACEF is a parallelogram; (2) When the size of ∠ B satisfies what condition, the quadrilateral ACEF is a diamond? Please prove your conclusion; (3) Is it possible that the quadrilateral ACEF is a rectangle? Why?
(1) It is proved that: ∵ ED is the vertical bisector of BC, EB = EC. 3 = ? ACB = 90 °, 2 and ∠ 4 are mutually complementary, 1 and ∠ 3 are mutually complementary, ? 1 = ∠ 2. ? AE = CE. Also ? AF = CE, ace and △ EFA are isosceles triangles. AF = AE, ? f = 5, ? FD ⊥ BC, AC ⊥ BC, ? AC
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