Question

12. A man lies on a road and observes that a temple and a flagstaff on it subtend equal angles at his eye. If the height of the temple be 10m and that of the flagstaff 20m, find the angle subtended by the temple at the man's eye and its distance from the man is ​

Answer 1

Answer:

sorry I don't know

Answer 2

Solution :-

given that,

→ AB = Height of tample = 10 m

→ DA = Height of flagstaff = 20 m

→ BC = Distance of tample from man = Let x m .

→ ∠DCA = ∠ACB = Let θ .

So, In ∆ABC ,

→ tan θ = AB/BC

→ tan θ = (10/x) ---------- Eqn.(1)

and, In ∆DBC ,

→ tan 2θ = DB/BC

→ tan 2θ = (30/x) ---------- Eqn.(2)

we know that,

→ tan 2θ = 2•tan θ / (1 - tan²θ)

→ (30/x) = (20/x) / {1 - (10/x)²}

→ (30/x) = (20/x) / {1 - (100/x²)}

→ (30/x) = (20/x) / {(x² - 100)/x²}

→ (30/x) = (20/x) * {x²/(x² - 100)}

→ (3/x) = 2x/(x² - 100)

→ 2x² = 3x² - 300

→ 3x² - 2x² = 300

→ x² = 300

→ x = 10√3 (Ans.)

therefore ,

→ tan θ = (10/x)

→ tan θ = 10/10√3

→ tan θ = 1/√3

→ tan θ = tan 30°

→ θ = 30° (Ans.)

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