On a straight line passing through the foot of a tower, two points C and D are at distance of 4cm and 16cm from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.
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Let AB be x (height of tower)
Let ∠C be ∅ and ∠D be (90-∅)
Given: CB = 4cm and BD = 16cm
Proof: In ΔABC,
tan∅ = x/4 → 1
In ΔABD,
tan(90-∅) = x/16 → 2
We know that tan(90-∅) = cot∅ and cot∅ = 1/tan∅
1=2 ⇒ tan∅ = 1/ tan
x/4 = 16/x ( From 1 and 2)
∴ x² = 16 × 4
⇒ x² = 64
⇒ x = √64
⇒ x = 8
∴ Height of tower is 8 cm
Hope this helps ^-^ !!!
Let ∠C be ∅ and ∠D be (90-∅)
Given: CB = 4cm and BD = 16cm
Proof: In ΔABC,
tan∅ = x/4 → 1
In ΔABD,
tan(90-∅) = x/16 → 2
We know that tan(90-∅) = cot∅ and cot∅ = 1/tan∅
1=2 ⇒ tan∅ = 1/ tan
x/4 = 16/x ( From 1 and 2)
∴ x² = 16 × 4
⇒ x² = 64
⇒ x = √64
⇒ x = 8
∴ Height of tower is 8 cm
Hope this helps ^-^ !!!
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height of the tower is 8 cm
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