Question

In the diagram, TC represents a vertical building. The points, A and B, are on the same level as the foot C of the building such that ATC = 40° and BTC = 56°. If BT is 29 m longer than AT, find

(a) the height of the building,
(b) the distance AB.

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Answer 1

(a) The height of the building is 60.06 m

(b) The distance AB is 139.43 m

Explanation:

The given parameters are

Given that segment BT = segment AT + 29

By trigonometric ratios, we have;

cos∠ATC = CT/AT

cos∠BTC = CT/BT

Therefore, we have;

cos(40°) = CT/AT.................................(1)

cos(56°) = CT/BT = CT/(AT + 29).....(2)

cos(56°) = CT/(AT + 29)......................(3)

From equation (1)

CT = AT×cos(40°)

From equation (3)

AT×cos(56°) + 29 × cos(56°) = CT

Therefore;

AT×cos(40°) = AT×cos(56°) + 29 × cos(56°)

AT×cos(40°) - AT×cos(56°) = 29 × cos(56°)

AT×(cos(40°) - cos(56°)) = 29 × cos(56°)

AT = 29 × cos(56°)/(cos(40°) - cos(56°)) = 78.4 m

TC = CT = AT×cos(40°) = 78.4×cos(40°) = 60.06 m

The height of the building = 60.06 m

(b) BT = AT + 29 = 78.4 m + 29 m= 107.4 m

AB = AT×sin(∠ATC ) + BT×sin(∠BTC) = 78.4×sin(40°) + 107.4×sin(56°) = 139.43 m

The distance AB = 139.43 m.