A practical application problem of solving triangles. . No, help.

Solution: Let the height of the tall tower be H, the height of the short tower be H, the elevation angle of the tall tower under the short tower be A, and the elevation angle of the tall tower at point O be B. 。

The elevation angle of one tower is twice that of the other tower, measured at the bottom of the two towers, so the elevation angle of the short tower under the high tower is a/2.

That is, tana=H/ 120, tan(a/2)=h/ 120,

According to the angle doubling formula, h/120 = [2 * (h/120)] [1-(h/120) 2] ①,

The elevation angle of the two towers is the complementary angle measured at the midpoint o of the connecting line between the towers, so the elevation angle of the low tower at the point o is π-b, that is, tanb=H/60 and tan(π-b)=h/60.

According to the inductive formula, H/60 = 60/H 2,

Simultaneous ① ② get H=90 and h=40.

That is, the height of the two towers is 40 90.

or

Solution: Let AB height XM and CD height ym have an elevation angle of A and the other elevation angle of 2a.

Judging from the meaning of the question.

Because of the similarity,

So 60/n=m/60

Because the elevation of one tower top is twice that of the other.

tan2a = m/ 120 = 2 tana/( 1-(tana)2)

Tana=n/ 120 n=40 is used to solve the equation.

m=90

Therefore, the heights of the two towers are 40 meters and 90 meters respectively.