Question

AB is a vertical pole with its foot B on the level ground. C is a point on AB such that AC : CB = 9 : 1. If the parts AC and CB subtent equal angles at a point on the ground which is at a distance of 20 metres from the foot of the pole, find the height of the pole.

Answer 1

Refer the attachment for figure.

Here, let AC = 9x and BC = x. Let angle subtended by them be 'a'.

Now,

AB = AC + BC

→ AB = 9x + x

→ AB = 10x

By Pythagoras Theorem,

CO² = x² + 20²

→ CO² = 400 + x²

And, AO² = (10x)² + 20²

→ AO = $\sqrt{(100x^2 + 400)}$

Now, we know that sin∅ = Perpendicular/Hypotenuse

and, cos∅ = Base/Hypotenuse

So, sin(a) = BC/CO

and, cos(a) = BO/CO

Now, in ∆ABO,

sin(2a) = AB/AO

But, we know that, sin2∅ = 2cos∅sin∅

→ sin(2a) = 2cos(a)sin(a)

→ AB/AO = 2 × BO/CO × BC/CO

→

$\frac{10x}{AO} = 2 \times \frac{20}{CO} \times \frac{x}{CO} \\ \\ \implies \: \frac{10x}{AO} = \frac{40x}{{CO}^{2} }$

Now, cancel x from both sides and out the values of AO and CO²

→

$\frac{10}{ \sqrt{100 {x}^{2} + 400 } } = \frac{40}{400 + {x}^{2} }$

Crops multiply,

$40( \!\sqrt{100 {x}^{2} + 400)} \!= \!10(400 + {x}^{2} ) \\ \\ 40( \!\sqrt{\!100 {x}^{2} + 400)} \!= \!(\!4000 + \!10{x}^{2} )$

Now square both the sides,

→ (40)²(100x² + 400) = (4000 + 10x²)²

→1600(100x² + 400) = (4000)² + (10x²)² + 2(4000)(10x²)

→ 1,60,000x² + 6,40,000 = 1,60,00,000 + 100x⁴ + 80,000x²

After Adding and subtracting like terms, we get

→ 80,000x² = 1,53,60,000 + 100x⁴

→ 100x⁴ - 80,000x² + 1,53,60,000 = 0

→ 100(x⁴ - 800x² + 153600) = 0

After Dividing by 100 on both sides, we get

→ x⁴ - 800x² + 153600 = 0

→ x⁴ - 480x² - 320x² + 153600 = 0

→ x²(x² - 480) - 320(x² - 480) = 0

→ (x² - 320)(x² - 480) = 0

Hence, x² = 320 or x² = 480

→ x = √320 or x = √480

→ x = 8√5 or x = 4√30

So, height of pole = AB = 10x

→ 10x = 80√5 or 40√30

Hence, your answer is 80√5 metre or 40√30 metre

Answer 2

According to the problem given,

AC : CB = 9:1

Let AC = 9x m,

CB = x m

BP = 20 m

And <BPC = <CPA = y°

Now,

i ) In ∆CBP,

tan y = CB/BP = x/20 ---( 1 )

ii) In ∆ABP ,

tan ( <BPC+<CPA)

= tan (2y) = AB/BP

=> (2tan y)/(1-tan²y) = 10x/20

=> (2x/20)/[1-(x/20)²]=x/2

[ from (1)]

=> (x/10) = (x/2)[1-x²/400]

=> (x/10 )(2/x) = 1 - x²/400

=> (1/5)-1 = -x²/400

=> (1 - 5 )/5 = -x²/400

=> -4/5 = -x²/400

=> (4×400)/5 = x²

=> 4×80 = x²

=> x = √4×80

=> x = 8√5

Therefore,

Height of the pole = AC+BC

= 9x+x

= 10x

= 10 × (8√5)

= 80√5 m

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