Question

angle of depression of top and bottom of building 50m high as observed from top of tower are 30° and 45° . find height of tower and distance between tower and building​

Answer 1

Answer:

The angles of depression of the top and bottom of a building 50 meters high as 4 observed from the top of a tower are 30 and 60, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

In ΔBTP,

= TP/BPBP = TP√3In ΔGTR,

In ΔBTP,tan 30° = TP/BP1/√3 = TP/BPBP = TP√3In ΔGTR,tan 60° = TR/GR

In ΔBTP,tan 30° = TP/BP1/√3 = TP/BPBP = TP√3In ΔGTR,tan 60° = TR/GR√3 = TR/GR

In ΔBTP,tan 30° = TP/BP1/√3 = TP/BPBP = TP√3In ΔGTR,tan 60° = TR/GR√3 = TR/GRGR = TR/√3

In ΔBTP,tan 30° = TP/BP1/√3 = TP/BPBP = TP√3In ΔGTR,tan 60° = TR/GR√3 = TR/GRGR = TR/√3As BP = GR

In ΔBTP,tan 30° = TP/BP1/√3 = TP/BPBP = TP√3In ΔGTR,tan 60° = TR/GR√3 = TR/GRGR = TR/√3As BP = GRTP√3 = TR/√3

In ΔBTP,tan 30° = TP/BP1/√3 = TP/BPBP = TP√3In ΔGTR,tan 60° = TR/GR√3 = TR/GRGR = TR/√3As BP = GRTP√3 = TR/√33 TP = TP + PR

In ΔBTP,tan 30° = TP/BP1/√3 = TP/BPBP = TP√3In ΔGTR,tan 60° = TR/GR√3 = TR/GRGR = TR/√3As BP = GRTP√3 = TR/√33 TP = TP + PR2 TP = BG

In ΔBTP,tan 30° = TP/BP1/√3 = TP/BPBP = TP√3In ΔGTR,tan 60° = TR/GR√3 = TR/GRGR = TR/√3As BP = GRTP√3 = TR/√33 TP = TP + PR2 TP = BGTP = 50/2 m = 25 m

In ΔBTP,tan 30° = TP/BP1/√3 = TP/BPBP = TP√3In ΔGTR,tan 60° = TR/GR√3 = TR/GRGR = TR/√3As BP = GRTP√3 = TR/√33 TP = TP + PR2 TP = BGTP = 50/2 m = 25 mNow, TR = TP + PR

TR = (25 + 50) m

TR = (25 + 50) mHeight of tower = TR = 75 m

TR = (25 + 50) mHeight of tower = TR = 75 mDistance between building and tower = GR =TR/√3

GR = 75/√3 m = 25√3 m

Step-by-step explanation:

Hope this helps you thanku ｡◕‿◕｡