**Problem 1 :**

Two ships are sailing in the sea on either side of the lighthouse. The angles of depression of two ships as observed from the top of the lighthouse are 60° and 45° respectively. If the distance between the ships is 200[(√3 + 1)/√3] metres, find the height of the lighthouse.

**Solution :**

In triangle ABD

tan 45 = DB/AB

1 = DB/AB

AB = DB ---------(1)

In triangle DBC,

tan 60 = DB/BC

√3 = DB/BC

BC = DB/√3 ---------(2)

(1) + (2)

AB + BC = 200[(√3+1)/√3]

DB + (DB/√3) = 200[(√3+1)/√3]

DB(1 + (1/√3)) = 200[(√3+1)/√3]

DB[(√3+1)/√3] = 200[(√3+1)/√3]

DB = 200

Hence the height of the light house is 200 m.

**Problem 2 :**

A building and a statue are in opposite side of a street from each other 35 m apart. From a point on the roof of building the angle of elevation of the top of statue is 24° and the angle of depression of base of the statue is 34° . Find the height of the statue. (tan 24° = 0.4452, tan 34° = 0.6745)

**Solution :**

In triangle AED,

tan 24 = ED/AD

0.4452 = ED/35

ED = 0.4452(35)

ED = 15.582

In triangle ABC,

tan 34 = AB/BC

0.6745 = AB/35

0.6745(35) = AB

AB = 23.60

AB = DC

Height of statue = 15.58 + 23.60

= 39.18 m

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