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Examples of solving triangles in mathematics
The positive solution is as follows ~ ~ remember to give points.

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.