Question

Q.There are two temples, one on each bank of a river, just opposite to each other. One temple is 50m high. From the top of this temple, the angles of depression of the top &the foot of the other temple are 300 & 600 respectively. Find the width of the river and the hight of the other temple.

Answer 1

Let AB and CD be the two two temples and Bc be the width of the river.

Then ∠ ADE =30⁰ and ∠ ACB = 60⁰

Given A = 50m

∴ In Δ ABC

tan 60⁰ = AB/BC

√3 = 50/BC

BC = 50/√3

= 50√3/3

= 28.867m

and In Δ AED

tan 30⁰ = AE/ED

tan 30⁰ = AB - BE/BC

tan 30⁰ = AB - DC/BC

 1/√3 = 50 - DC /50/√3

50/3 = 50 - DC

DC = 50 - 50/3

= 150/3 - 50/3

= 100/3 m

= 33.33 m

Hence the width of the river is 28.867m and height of the other temple is 33.33m.

Answer 2

Answer:

50/√3 m, 100/3 m

Step-by-step explanation:

Let the width of the river be  d m.

Given that height of the temple AB (say) is 50 m. Assume the height of the other temple CD (say) x m; i.e, CD = x m.

Assume a point M on AB such that MD || AC.

Now it is given that angle of depression from top of the temple of the length (50m) to  C and D are 30 and 60 degrees respectively.

Consider the triangle Δ BAC,  then:

$tan (30^o)=\frac{d}{50} \\\implies d = 50/\sqrt{3} m.....(1)$

Consider the triangle Δ BMD, then :

$tan(60^o) =\frac{50/\sqrt{3} }{MB} \\\implies MB = 50/3 \,m$

Height of the other temple is CD = AB - MB = 50 - 50/3 = 100/3m $\approx$ 33.3m

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