High School Math Solving Triangles Question 2: How to do it?

Solution: (1) From a, b, c into a series of equal differences, and a+b+c= to obtain b= 3, so there is a+c=2 3

From 2b = 3ac we get 2sin b = 3 sinasinc = 3 2, so sinasinc = 1 2

So cos(a+c)=cosacosc sinasinc=cosacosc 1 2

i.e. cosacosc 1 2 = 1 2, we can get cosacosc=0 so cosa=0 or cosc=0, i.e. a is a right angle or c is a right angle.

So a = 2, or a = 6

2) When A = 2, B = 3, C = 6 in RT ABC, A = 1, B = 3 2, C = 1 2S (ABC) = 1 2 BC = 3 8

When a = 6, b = 3, c = 2

In RT ABC, a = 1, b = 3, c = 2s ( abc) = 1 2 ab = 3 2

Satisfied, thank you!

tan30° = how much I forgot, and then this can calculate the length of b, and the area can be calculated according to this.

a= 2 or 6, abc is a right triangle, just discuss it in different situations.

The first question refers to the formula of using the sum of laughter and confusion in the two corners.

The second question makes use of the cosine theorem.

Because b=2a, sinb=2sina (sinusoidal theorem).

sin(a+60)=2sina

The solution is a = 30°

(2)……Pick up your answers.

The range of u=a+ 6 is (6,2 3), and the range of v=sinu is (1 2,1], so the range of a+b=2 3v is (3,2 3], so the range of a+b+c is (2 3,3 3).

3)s△abc=(1/2)absinc=(1/2)*2sina*2sinb*√3/2=√3sinasinb

-√3/2][cos(a+b)-cos(a-b)]=(√3/2）[1/2+cos(a-b）],a-b|<π/2，cos|a-b|The range of is (0,1], so the range of s abc is ( 3 4,3 3 4).