Math problems in grade three: right triangle with elevation angle and depression angle.

(1) solution: let c be the continuation of CE⊥AB at point e, and let c' be C'F⊥AB at point f.

According to the question, ∠ cab = 30, ∠ CBE = 60. ∴∠ACB=30 .

∴∠cab=∠acb=30,∴ab=cb= 100m。

In RT△CEB, ∠ CEB = 90, ∠ CBE = 60. ∴ ce = CB× sin60 =100× √ 3/2 = 50 √ 3 (m)

∫CC '//AB .∴C'F=CE=50√3m≈86.6m (the distance between parallel lines is equal everywhere).

The height of the balloon is 86.6 meters.

(2) Solution: We can get ∠ c 'af = 45 and ∠ c 'fa = 90. ∴∠AC'F=45 .

∴∠C'AF=∠AC'F=45 ,∴AF=C'F=50√3m。

∠∠CEB = 90，∠CBE=60 .∴∠BCE=30 .

In RT△CEB, ∠ CEB = 90, ∠ BCE = 30. ∴BE=CB/2=50m。

∴fe=ae-af=ab+be-af= 100+50-50√3=( 150-50√3)(m)

∠∠AFC ' =∠BEC = 90 .∴CC'=FE=( 150-50√3)(m)

∴ Balloon speed v = (150-50 √ 3) ÷10 = (15-5 √ 3) ÷ 6.34 (m/s).