a girl of height 90cm is walking away from the base of a lamp-post at a speeded of 1.2m/s of the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
(chapter - triangles.)
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Hi!
Here is the answer to your question.

Let the girl be at point D on the ground from the lamppost after 4 sec.
∴ AD = 1.2 m/sec × 4 sec = 4.8 m = 480 cm
Suppose the length of the shadow of the girl be x cm when she at position D.
∴ CD = x cm
In ∆CDE and ∆CAB
∠CDE = ∠CAB (90°)
∠DCE = ∠ACB (Common)
∴ ∆CDE ∼ ∆CAB (AA similarity)

∴ Length of her shadow after 4 second is 160 cm.
Here is the answer to your question.

Let the girl be at point D on the ground from the lamppost after 4 sec.
∴ AD = 1.2 m/sec × 4 sec = 4.8 m = 480 cm
Suppose the length of the shadow of the girl be x cm when she at position D.
∴ CD = x cm
In ∆CDE and ∆CAB
∠CDE = ∠CAB (90°)
∠DCE = ∠ACB (Common)
∴ ∆CDE ∼ ∆CAB (AA similarity)

∴ Length of her shadow after 4 second is 160 cm.
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