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In the triangle ABC, ab = 3, AC = 4, D is the midpoint of BC edge, ad = (√ 37) / 2, find the angles a and BC
Extend ad to e so that ad = De
According to cosine theorem, cosace = (AC 2 + EC? - AE 2) / 2Ac × EC
= -1/2
∠ACE=120°
∠BAC=60°
BC²=AB²+AC²-2AC×AB cocBAC=9+16-12=13
BC=√13
In △ ABC, BC = 3, ab = 2, and sinc sinB=2 5（ 6 + 1), then a=______ .
∵ AB = 2 and sinc
sinB=2
5（
6+1），
∴2
AC=2
5（
6+1），
∴AC=
6-1，
∴cosA=4+(
6−1)2−9
2×2×(
6−1)=-1
2，
∴A=120°，
So the answer is: 120 degrees
In △ ABC, if Sina: SINB: sinc = 3:5:7, then the degree of the maximum inner angle of the triangle is equal to?
By using the sine theorem: A / Sina = B / SINB = C / sinc ∵ Sina: SINB: sinc = 3:5:7 ᙽ A: B: C = 3:5:7, let a = 3T, B = 5T, C = 7T ∵ big side to large angle, ? C maximum use cosine theorem cos C = (a ∧ B ? C ? Sina: SINB: sinc = 3:5:7 ? A: B: C = 3:5:7 ? A: B: C = 3:5
In the triangle ABC, given the vector ab × AC = 2, s triangle ABC = 2 (1), find the value of Tana (2) if SINB = 2cosasinc, find the length of BC
(1) S triangle ABC = AB * acsina / 2 = 2
Vector ab × AC = AB * ac * cosa = 2
So Tana = 2 * s triangle ABC / (vector ab × AC) = 2
（2）sinB=sin(A+C)=sinAcosC+cosAsinC
So 2cosasinc = sinacosc + cosasinc
So sinacosc cosasinc = sin (A-C) = 0
So SINB = 2cosasinc = 2cosasina = sin2a
So B = 2A
Because a + B + C = 180 degrees
So a = C = 45 ° and B = 90 °
Because the vector ab × AC = AB * ac * cosa = the square of AB = 2
So BC = AB = radical 2
It is known that the triangle ABC is in the first quadrant, the angle a (1.1) (5,1) a is 60 ° and the angle B is 45 °, AC.BC Find the line of AC and BC and y The distance between the intersection points of B (5,1) sorry, it's too urgent
AB：y=1
AC：y=3^(1/2)*x+1-3^(1/2)
BC：y=-x+6
Distance = 6 - (1-3 ^ (1 / 2)) = 5 + 3 ^ (1 / 2)
Is a mathematical problem: known in the first quadrant of the triangle ABC, a (1,1), B (5,1), angle a is equal to 60 degrees, angle B is equal to 40
In the triangle ABC of the first quadrant, a (1,1), B (5,1), the angle a is equal to 60 degrees, and the angle B is equal to 45 degrees. (1) find the equation of AB side; (2) the equation of the straight line where AC and BC are 1. K (AB) = (1-1) / (5-1) = 0, then the straight line AB is y = 12
In the triangle ABC in the first quadrant, a (1,1) B (5,1) angle a is 60 degrees and angle B is 45 degrees Problem solving: in the triangle ABC in the first quadrant, a (1,1) B (5,1) angle a is 60 degrees and angle B is 45 degrees. 1. Find the equation of AC side 2. Find the length of the height on the edge ab of the triangle ABC
AB is parallel to the X axis, angle A60, the first quadrant of ABC, the slope of AC = Tan 60 = sqrt (3) = 1.732
Point slant equation: Y-1 = 1.732 (x-1)
Similarly, BC equation: Y-1 = - Tan 45 (X-5) = 5-x (slope
It is known that in △ ABC, the angle a = 60 degrees, s △ ABC = √ 3, B + C = 5, find the edge of A I do B = 1 or 4, C = 1 or 4, and then find a, there are four answers. What's going on?
S=bcsinA/2=√3
∴bcsin60°/2=√3
∴bc=4
B + C = 5, be careful not to solve the value of B and C
∴a²=b²+c²-2bcsinA
=b²+c²-bc
=(b+c)²-3bc=25-12=13
ν a = √ 13 A = - √ 13, omit
Given that the triangle ABC is in the first quadrant, if a (1,1), B (5,1), the angle a is equal to 60 degrees, and the angle B is equal to 45 degrees, find: (1) the equation of the line where the edge AB lies
0
0
According to CoSb / COSC = - B / 2A + C
B = 120 ° is obtained
According to s = 1 / 2acsinb = 5, root number 3
C = 5
According to CoSb = (a ^ 2 + C ^ 2-B ^ 2) / 2Ac = - 1 / 2
It can be solved that B = root 61
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