**Problems Involving Angle of Elevation and Depression :**

Here we are going to see, some practice problems involving angle of elevation and depression.

**Question 1 :**

From the top of a tree of height 13 m the angle of elevation and depression of the top and bottom of another tree are 45° and 30° respectively. Find the height of the second tree. (√3 = 1.732)

**Solution :**

AB = 13 m = DC

In triangle AED,

tan 45 = ED/AD

1 = ED/AD

AD = ED --(1)

In triangle ABC,

tan 30 = AB/BC

1/√3 = 13/BC

BC = 13√3 ----(2)

AD = BC

By applying the value of AD in (1), we get

ED = 13√3

height of second tree = ED + DC

= 13√3 + 13

= 13(√3 + 1)

= 13(1.732 + 1)

= 13(2.732)

= 35.516 m

Hence the height of the second tree is 35.52 m

**Question 2 :**

A man is standing on the deck of a ship, which is 40 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30° . Calculate the distance of the hill from the ship and the height of the hill. (√3 = 1.732)

**Solution :**

In triangle AED,

tan 60 = ED/AD

√3 = ED/AD

AD = ED/√3 -----(1)

In triangle ABC,

tan 30 = AB/BC

1/√3 = 40/BC

BC = 40√3 -----(2)

BC = AD

(1) = (2)

40√3 = ED/√3

ED = 40√3(√3)

= 40(3)

ED = 120 m

Height of hill = ED + DC

= 120 + 40

= 160 m

Distance from hill to ship = 40√3

= 40(1.732)

= 69.28 m

**Question 3 :**

If the angle of elevation of a cloud from a point ‘h’ meters above a lake is θ_{1} and the angle of depression of its reflection in the lake is θ_{2} . Prove that the height that the cloud is located from the ground is h(tan θ_{1}+ tan^{ }θ_{2})/(tan θ_{2 }- tan^{ }θ_{1})

**Solution :**

In triangle AED,

tan θ_{1 }= ED/AD

tan θ_{1 }= x/AD

AD = x/tan θ_{1}

AD = x cot θ_{1 } -----(1)

In triangle ADC,

tan θ_{2 }= DC/AD

tan θ_{2 }= (2h + x)/AD

AD = (2h + x)/tan θ_{2}

AD = (2h + x)cot θ_{2} -----(2)

(1) = (2)

x cot θ_{1 = }(2h + x)cot θ_{2}

x cot θ_{1} = 2h cot θ_{2} + x cot θ_{2}

x (cot θ_{1}^{ }- cot θ_{2}) = 2h cot θ_{2}

x (tan θ_{2}^{ }- tan θ_{1})/(tan θ_{1}^{ }tan θ_{2}) = 2h (1/tan θ_{2})

2h = [xtan θ_{2}(tan θ_{2}^{ }- tan θ_{1})]/(tan θ_{1}^{ }tan θ_{2})

2h = x(tan θ_{2}^{ }- tan θ_{1})/tan θ_{1}

x = 2h tan θ_{1}/(tan θ_{2}^{ }- tan θ_{1})

height of cloud = x + h

= [2h tan θ_{1}/(tan θ_{2}^{ }- tan θ_{1})] + h

= [2h tan θ_{1}+ h(tan θ_{2}^{ }- tan θ_{1})]/(tan θ_{2}^{ }- tan θ_{1})]

= h[tan θ_{1}+ tan θ_{2}]/(tan θ_{2}^{ }- tan θ_{1})]

Hence proved.

After having gone through the stuff given above, we hope that the students would have understood, "problems involving Angle of Elevation and Depression".

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