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In the triangle ABC, a = 120 °, ab = 5, BC = 7, then the value of sina / SINB
Sine theorem: sinc = Sina * AB / BC
A=120°,AB=5,BC=7
∴sinC=5√3/14,cosC=11/14
sinB=sin（A+C）
=sinAcosC+sinCcosA
=3√3/14
∴sinA/sinB=7/3
Given that the three sides of a triangle satisfy the condition A2 − (B − C) 2BC = 1, then ∠ a=______ ．
∵ A2 − (B − C) 2BC = 1, ∵ a2 - (B-C) 2 = BC, which is reduced to B2 + c2-a2 = BC. From the cosine theorem, it is obtained that cosa = B2 + C2 − a22bc = 12, ∵ A is the inner angle of the triangle, ∵ a = 60 °
In the triangle ABC, the side lengths of angles a, B and C are a, B and C in turn. A, B and C satisfy the condition b square + C square - BC = C of a square and B
In the triangle ABC, the side lengths of angles a, B and C are a, B and C in turn. A, B and C satisfy the condition that b square + C square - BC = a square and C of B = 1 / 2 + radical 3
Find the value of angle a?
Find the value of tanb?
According to the cosine theorem: B2 + c2-2bccosa = A2, cosa = 1 / 2 and angle a = 60 ° are deduced. Then, by dividing both sides of B2 + C2 BC = A2 by B2 at the same time, a / b = radical 15 / 2 is deduced. From S = 1 / 2bcsina = 1 / 2acsinb, SINB = (B / a) Sina = 5 / 5, tanb = 1 / 2. (angle B must be an acute angle)
As shown in Figure 1, in the triangle ABC, the intersection of high AD and high be is h, and BH = AC, then the angle ABC =?
Is it OK to type slowly when ∵ - ahe + ∵ hae = 90 °, C + ∵ hae = 90 °?
It is known that the vertex of equilateral triangle DEF is on the edge of equilateral △ ABC to prove ad = be = CF
This is a good proof. You can understand it by drawing the dotted line
In tetrahedral oabc, if edges OA, OB and OC are perpendicular to each other, and OA = 1, OB = 2, OC = 3, G is the center of gravity of triangle ABC, what is the vector og
A (0,0,1), B (2,0,0), C (0,3,0) so Ba = (- 2,0,0), BC = (- 2,3,0)
BG=1/3（BC+BA）=（-3/4,1,1/3)
So og = ob + BG = (2 / 3,1,1 / 3) = 1 / 3 · √ 14
It is known that in △ ABC, BD bisects ∠ ABC, ed ‖ BC, EF ‖ AC, and it is proved that be = CF
∵ ed ∥ BC, EF ∥ AC, efcd are parallelogram, ∵ ed = CF, ∵ BD bisection ∥ ABC, ∥ EBD = ∥ FBD, ed ∥ BC, ∥ EDB = ∥ FBD, ∥ EBD = ∥ EDB, ∥ EB = ed, ∥ EB = CF
It is known that in the triangular pyramid s-abc, ∠ ACB = 90 ° and SA ⊥ plane ABC, ad ⊥ SC is in D. It is proved that ad ⊥ plane SBC
It is proved that: ∵ SA ⊥ plane ABC, ∵ BC ⊥ Sa; ∵ ACB = 90 °, that is, AC ⊥ BC, and AC and SA are two intersecting lines in plane sac, ∵ BC ⊥ plane sac; ad ⊂ plane sac, ∵ BC ⊥ ad, and ∵ SC ⊥ ad, and BC and SC are two intersecting lines in plane SBC, ≁ ad ⊥ plane SBC
In tetrahedral p-abc, PC is perpendicular to plane ABC, ab = BC = CA = PC, and the cosine value of dihedral angle b-ap-c is obtained
Let AB = BC = CA = PC = a
The plane PAC is perpendicular to the plane ABC
Take the midpoint D of AC to connect BD and PD
Know that BD is perpendicular to AC
So D is the projection of point B in plane PAC, and triangular pad is the projection of triangular PAB in plane PAC
The area of triangular pad is: S = (1 / 2) * 1 * 1 * (1 / 2) a ^ 2 = (a ^ 2) / 4. (1)
In triangle PAB, PA = Pb = (radical 2) a, ab = a
According to the cosine theorem, the cos angle APB = [2 + 2-1] / [2 * 2] = 3 / 4 is obtained
So sin angle APB = (root 7) / 4
The area of triangle PAB a = (a ^ 2) (1 / 2) * (root 2) (root 2) * (root 7) / 4 (2) can be obtained
=(a ^ 2) (root 7) / 4
Let the dihedral angle b-ap-c be α,
From the projection theorem: a * cos α = s
So: cos α = s / a = (1 / 4) / [(radical 7 / 4] = 1 / radical 7 = (radical 7) / 7
Are you satisfied with the above answers?
As shown in the figure, in tetrahedral p-abc, PC is perpendicular to plane ABC, ab = BC = CA = PC, then the cosine of dihedral angle b-pa-c is?
It's very easy to draw, to build a department, to do
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