Question

Cho tam giác ABC, góc A = 90 độ, đường cao AH = \(\sqrt{3}\),BH = 1 cm. Tím HC và góc ACB

Answer 1

Answer 2

Ta có: `AH^2 = BH . CH`.

`<=> sqrt 3^2 = 1 . CH`

`=> CH = 3 cm`.

`+, AC^2 = HC xx BC`.

`=> AC = sqrt (HC xx BC) = sqrt (3 xx 4) = sqrt 12`.

 

`=> AB = sqrt (HB xx BC) = sqrt (1 xx 4) = 2`.

`=> sin(x) = (AB)/(BC)`

`=> sin(x) = 1/2`

`=> x = 30^o`.