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As shown in the figure, in △ ABC, the angle a = 60 ° and the circle O with BC as diameter intersects with AB and AC at points D and e respectively. Try to judge the shape of △ deo and prove it
Delta deo is an equilateral triangle
Proof: because od = OE
So △ deo is an isosceles triangle
Connect CD, BC is diameter
So ∠ BDC = 90 degrees
Because ∠ a = 60 degrees
Therefore, DCE = 30 degrees
Because ∠ DOE is the center angle of arc De
So ∠ DOE = 60 degrees
Because the triangle deo is isosceles triangle
So the triangle deo is a regular triangle
As shown in the figure, a circle with diameter BC intersects with two sides AB and AC of the triangle at two points D and e respectively, and BD = EC. It is proved that the triangle ABC is an isosceles triangle
Certification:
∵BD=EC
Ψ arc BD = arc EC
Arc BDE = arc BD + Arc de, arc CED = arc EC + arc De
Ψ arc BDE = arc CED
∴∠C=∠B
The triangle ABC is an isosceles triangle
As shown in the figure, in △ ABC, ab = AC, D is a point on the edge BC, AB and BD are the adjacent edges, making ▱ ABDE, connecting AD and EC. (1) prove: △ ADC ≌ △ ECD; (2) if BD = CD, prove: quadrilateral adce is a rectangle
It is proved that: (1) the ∵ quadrilateral ABDE is a parallelogram (known), ∵ ab ∥ De, ab = de (the opposite sides of the parallelogram are parallel and equal); ∵ B = ∠ EDC (two lines are parallel and the same angles are equal); ∵ AB = AC (known), ∵ AC = de (equal substitution), ∵ B = ∠ ACB (equal angle), ∵ EDC = ∠ ACD (equal substitution); ∵ in △ ADC and △ ECD, AC = ed In △ ABC, we can find that ∠ ACD = ∠ edcdc = CD (common edge), ≌ △ ADC ≌ △ ECD (SAS); (2) ∵ quadrilateral ABDE is a parallelogram (known), ∵ BD ∥ AE, BD = AE (opposite sides of parallelogram are parallel and equal), ∥ AE ∥ CD; and ∵ BD = CD, ≌ AE = CD (equivalent substitution), ≌; quadrilateral adce is a parallelogram (opposite sides are parallel and equal quadrilateral is a parallelogram); in △ ABC, the, AB = AC, BD = CD, ⊥ ad ⊥ BC (three in one property of isosceles triangle), ⊥ ADC = 90 ° and ▱ adce is rectangle
As shown in the figure, in △ ABC, ab = AC, D is the midpoint of BC, and the quadrilateral ABDE is a parallelogram
Because it is a parallelogram, AE = BD and AE / / BD, and because D is the midpoint of BC, BD = DC, AE = DC and AE / / DC, AECD is a parallelogram
Because AB = AC, D is the midpoint of BC, so the angle ADC is 90 degrees, so the parallelogram AECD is rectangular
In △ ABC, a, B and C are three internal angles, Tan C = - (COS a-cos b) / (sin a-SiN b), sin (B-A) = cos C, and the values of a and C are obtained
tan C=-(cos A-cos B)/(sin A-sin B）
sinC/cosC=-(cos A-cos B)/(sin A-sin B）
sinAsinC-sinBsinC=cosBcosC-cosAcosC
sinAsinC+cosAcosC=cosBcosC+sinBsinC
cos(A-C)=cos(B-C)
A-c = B-C, or a-c = C-B
If a-c = B-C, = = > A = B does not conform to the meaning of the question (denominator is 0)
∴A-C=C-B==>A+B=2C,
A+B+C=180º ,3C=180º,C=60º
sin(B-A)=cos C=1/2
∴B-A=30º (∵-120º≤B-A≤120º)
B-A=30º ,B+A=120º ,∴B=75º,A=45º
∴A=45º,C=60º
,
In the triangle ABC, ab = C, BC = a, AC = B, a, B is the solution a of X square-5x + 6
3. Find the two roots of the area equation x - 5x + 6 = 0 of triangle ABC as x = 2 or 3, because a
In triangle ABC, ab = 3, AC = 2 √ 2, cosa is a root of the equation 3x square + 5x-2 = 0. 1, find the area s of triangle ABC; 2, find cos (2B + 2C)
First solve the quadratic equation of one variable, notice 1 > cosa > - 1, so find cosa = 1 / 3
Then we can find Sina = 2 root 2 / 3
The area is s = (1 / 2) bcsina;
cos2(B+C)=2[cos(B+C)]^2-1=2cosAcosA-1
In △ ABC, if 3sina + 4cosb = 6 and 4sinb + 3cosa = 1, then ∠ C equals ()
A. 30 ° B, 150 ° C, 30 ° or 150 ° D, 60 ° or 120 ° C
I will square the above two equations respectively and add them together to get sinc = sin (a + b) = 1 / 2, ∧ C = 30 ° or 150 ° and choose C. but the answer is d. why should I leave ∧ C = 150?
Because: 3sina + 4cosb = 6 (1)
3cosA+4sinB=1 (2)
So the square of (1) is 9sina * Sina + 16cosb * CoSb + 24sinacosb = 36 (3)
(2) The square of is 9cosb * CoSb + 16sina * Sina + 24sinacosb = 1 (4)
So (3) + (4) 9sina * Sina + 9cosa * cosa + 16sinb * SINB + 16cosb * CoSb + 24sinacosb + 24cosasinb = 9 + 16 + 24sin (a + b) = 36 + 1 = 37
So 9 + 16 + 24sin (a + b) = 37
So sin (a + b) = 1 / 2
So a + B = 30 degrees or 150 degrees
We can judge that it should be
If a + B = 30 degrees
cosA>√3/2
3cosA>3√3/2>1
3cosa + 4sinb = 1 is not consistent
So only
A + B = 150 degrees
Therefore, C = 30 degrees
A
In △ ABC, 3sina + 4cosb = 6, 4sinb + 3cosa = 1, then the value of ∠ C is______ ．
If C = 56 π, then 0 ＜ a ＜ π 6, thus Cosa ＞ 32, 3cosa ＞ 1 is contradictory to 4sinb + 3cosa = 1 (because 4sinb ＞ 0 is constant), so C = π 6. So the answer is: π 6
In the triangle ABC, if the angle c = 90 ° then the value of the function y = Sina ^ 2 + 2sinb has a maximum and a minimum,
In a right triangle, a and B are complementary to each other,
sinB=cosA
y=sinA^2+2cosA
=1-cosA^2+2cosA
=-(cosA-1)^2+2
A is between (0 ° and 90 °), so the maximum and minimum values cannot be obtained
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