A 1.3 metre tall boy is standing at some distance from a 40 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° 60° as he walks towards the building. Find the distance ,correct upto 2 decimal places ,that he walked towards the building.
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Given A 1.3 metre tall boy is standing at some distance from a 40 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° 60° as he walks towards the building. Find the distance ,correct up to 2 decimal places ,that he walked towards the building.
- Boy is 1.3 m tall, PQ = 1.3 m
- Now building is 40 m tall
- So AB = 40 m
- So angle APC = 30 degree
- Boy moves from Q to Ranges to 60 degree
- Angle of elevation changes to 60 degree.
- Therefore angle ASC = 60 degree
- So PQ = CB = 1.3 m
- So AC = AB – CB
- = 40 – 1.5
- AC = 38.5 m
- Since tower is vertical, angle ACP = 90 degree.
- In right angle triangle APC,
- So tan P = AC / PC
- So tan 30 = 38.5 / PC
- 1/√3 = 38.5 / PC
- So PC = 38.5 √3
- Now tan S = AC / SC
- So tan 60 = 38.5 / SC
- So √3 = 38.5 / SC
- So SC = 38.5 / √3
- Now PC = PS + SC
- 38.5 √3 = PS + 38.5 /√3
- 38.5 (√3 – 1/√3) = PS
- So PS = 38.5 (3 – 1 / √3)
- So PS = 38.5 x 2 / √3
- = 77 / √3 x √3 / √3
- = 77 √3 / 3
- PS = 25.6√3 m
- Since QR = PS, QR = 44.44 m
Therefore he walked 44.44 m towards the building.
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