**Trigonometry Practical Problems Using Angle of Elevation :**

Here we are going to see, some example problems based on angle of elevation.

In order to solve word problems, first draw the picture to represent the given situation.

To find the questions 1 and 2, please visit the page "Angle of Elevation Practice Problems".

To find the questions 3 and 5, please visit the page "Questions Based on Angle of Elevation".

**Question 6 :**

The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?

**Solution :**

Let DC = x, ED = 15 - x

Height of the electric pole = AB = DC

AD = BC

In triangle AED,

tan θ = Opposite side / Adjacent side

tan 30 = ED/AD

1/√3 = (15 - x)/AD

AD = (15 - x)√3 ----(1)

In triangle BCE,

tan 60 = EC/BC

√3 = 15/BC

BC = 15/√3

BC = (15/√3) ⋅ (√3/√3)

BC = 15√3/3 = 5√3----(2)

(1) = (2)

(15 - x)√3 = 5√3

15√3 - x√3 = 5√3

15√3 - 5√3 = x√3

x = 10√3/√3

x = 10

Hence the height of the electric pole is 10 m.

**Question 7 :**

A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?

**Solution : **

By using angle bisector theorem,

Let the the two parts subtend equal angles at point A such that

CAB = DAC = θ

BC/DC = AB/AD

1/9 = 25/AD

AD = 25(9)

AD^{2} = AB^{2} + BC^{2}

225^{2} = 25^{2} + (10x)^{2}

100x^{2} = 225^{2} - 25^{2}

100x^{2 }= (225 + 25) (225 - 25)

100x^{2 }= (250) (200)

x^{2} = 500

x = 100√5 m

**Question 8 :**

A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8° . What is the height of the peak if the distance between consecutive milestones is 1 mile. (tan 4° = 0.0699, tan 8° = 0.1405)

**Solution :**

Let BC = x

In triangle BDC,

tan 8 = DC/BC

0.1405 = DC/x

DC = 0.1405 x ----(1)

In triangle ADC,

tan 4 = DC/AC

0.0699 = DC/(1 + x)

DC = 0.0699(1 + x) ----(2)

(1) = (2)

0.1405 x = 0.0699(1 + x)

0.1405 x = 0.0699 + 0.0699 x

(0.1405 - 0.0699)x = 0.0699

0.0706 x = 0.0699

x = 0.0699/0.0706

x = 699/706

x = 0.99

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

DC = 0.1405 (0.99)

DC = 0.14 mile

Hence the height of peak is 0.14 mile.

After having gone through the stuff given above, we hope that the students would have understood, "Trigonometry Practical Problems Using Angle of Elevation".

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