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In triangle ABC, it is known that 2 vector AB city vector AC = absolute value vector AB city absolute value vector AC setting angle CAB =& Find the value of the angle & and if cos (@~&)=4/3 of 7, where@belongs to (3/3/5 of 6) Find the value of cos@ In triangle ABC, it is known that 2 vectors AB city vector AC = absolute value vector AB city absolute value vector AC setting angle CAB =& Find the value of the angle & and if cos (@~&)=4/3 of 7, where@belongs to (3/3/5 of 6) Find the value of cos@
(1) Cos angle &=1/2, angle &=π/3
(2) Cos angle &=1/2, sin angle &=√3/2
Cos (@~&)= cos@cos &+ sin@sin &= cos@/2+ sin@3/2=4√3/7
(Sin@)^2+(cos@)^2=1
Joint solution to cos@
In triangle ABC, D is a point on the edge of BC, AD is vertical AB, vector BC=√3*BD, AD absolute value=1, find AC*AD
AD is vertical AB, so vector AB*AD=0.
Vector AC*AD=(AB+BC)*AD
=AB*AD+BC*AD
=0+BC*AD
=√3 BD*AD
=√3(BA+AD)*AD
=√3(BA*AD + AD*AD)
=√3(0+ AD*AD)
=√3 AD2
=√3.
In the triangle ABC, the vector AB·vector AC = the square of the absolute value of vector AC, and the shape of the triangle ABC is judged In the triangle ABC, the vector AB·vector AC = square of the absolute value of vector AC, and the shape of the triangle ABC is In the triangle ABC, the vector AB·vector AC=square of absolute value of vector AC, and the shape of the triangle ABC is judged
0
In △ABC, vector AB·vector AC=vector BA·BC=1, absolute value vector AB equals
0
Given the position of abc as shown in the figure, try to simplify the absolute value of root number (a square)-a-b + absolute value of c-a + square of root number (b-c) From small to large, ab0c
By known: a <0, a-b <0,c-a>0, b-c <0
Principle=|a|-|a-b||c-a||b-c|
=-A+(a-b)+(c-a)-(b-c)
=-A+a-b+c-a-b+c
=-A-2b+2c.
In triangle ABC, it is known that 2 times vector AB * vector AC = root number 3 absolute value vector AB * vector AC =3 vector BC square, find angle
2|Vector AB Vector AC|cosA=Root 3|AB||Vector AC|=3a^2===> cosA=Root 3/2, A=30°,
3A^2= cbRoot=3b^2+3c^2-6bc*Root3/2,3b^2-4cbRoot3+3c^2=0
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