1. In the triangle ABC, AB=2 root 3, B= pie /6, and area = root 3 to find the length of AC.
Solution: S=( 1/2)ac*sinB
So √ 3 = (1/2) * 2 √ 3 * (1/2) * a.
a=2
b^2=a^2+c^2-2ac*cosB
b^2=4+ 12-2*2*2√3*√3/2=4
b=2
That is, AC=2.
2. In the triangle ABC, logc-loga =-logsinb = logroot 2, and b is an acute angle. Try to judge the shape of triangle ABC.
Solution: -lg sinB=lg √2
So sinb = (√ 2) (- 1) = √ 2/2.
B is an acute angle, so b = 45.
lgc-lga=lg√2
lg(c/a)=lg√2
c/a=√2
sinC=√2sinA
Because sinc = sin (a+b) = √ 2/2sina+√ 2/2cosa.
So √2/2 Sina =√2/2 Xhosa.
Sina =cosA
A=45
So it is an isosceles right triangle.
3. In the triangle ABC, if b 2sin 2c = c 2sin 2b = bccosbcosc, try to judge the shape of the triangle ABC.
Solution: b 2sin 2c = c 2sin 2b.
So bsinC=csinB
b^2sin^2c=bccosBcosC
bcsinBsinC=bccosBcosC
sinBsinC=cosBcosC
cos(B+C)=0
B+C=90
This is a right triangle.
4. In quadrilateral ABCD, B = D = 90, A = 60, AB = 4°, AD = 5°. Find the length of AC and the value of BD/CD.
Solution: connect AC and BD, let ∠BAC=X, then ∠ CAD = 60-X.
AC=AB/cosX=AD/cos(60-x)
4cos(60-X)=5cosX
2cosX+2√3sinX=5cosX
2√3sinX=3cosX
tanX=√3/2
So BC=ABtanX=2√3.
In △ABC, using Pythagorean theorem, AC=2√7.
△ACD, Pythagorean theorem is used, CD=√3.
In △BCD, solve the triangle, BD 2 = BC 2+CD 2-2bc * CD COS120.
BD^2= 12+3+6=2 1
BD=√2 1
So BD/CD=√7.
5. In the triangle ABC, b=4 c=3 BC, the center line m= root number 37/2, and find the areas of A, A and triangle ABC.
Solution: Let the midpoint of BC be D, extend AD to E, make AD = DE, and connect CE.
AD=DE，BD=CD，∠CDE=∠BDA
△ABD and△△ ECD are congruent.
CE=3
In △ACE, cos ∠ CAE = (AC 2+AE 2-CE 2)/(2ac * AE) =11(2 √ 37).
In △ACD, Cd2 = Ad2+AC2-2ac * ADCOS ∠ CAE =13/4.
CD=√ 13/2
So a=√ 13.
In △ABC, COSA = (9+16-13)/(2 * 3 * 4) =1/2.
So A=60
s =( 1/2)bcsinA =( 1/2)* 3 * 4 *(3/2)= 3√3
6.AD is the bisector of triangle ABC, and it is known that AC = 2ab = 3a = 60. Find the length of AD.
Solution: in △ABC, BC 2 = AC 2+AB 2-2AC * ABCOSA = 7.
BC=√7
On d, DE in e is perpendicular to AC, and DF in f is perpendicular to AB.
Because AD is an angular bisector, DE=DF.
Let DE=x, then DF=x and AD=2x.
AE=AF=√3x
CE=2-√3x，BF=3-√3x
So CD 2 = 4x 2-4 √ 3x+4, BD 2 = 4x 2-6 √ 3x+9.
CD=BC-BD
Solve the equation to get x=3√3/5.
So AD=6√3/5.
I complicated the last question ~ ~ I consulted an expert, and it is much easier to do it with the area relationship.