In the previous section we saw how heights and distances can be determined using trigonometry. We also saw some solved examples. In this section we will see a few more solved examples.
Solved example 30.29
A 1.75 metre tall man, standing at the foot of a tower, sees the top of a hill 40 metres away at an elevation of 60o. Climbing to the top of the tower, he sees it at an elevation of 50o. Calculate the heights of the tower and the hill.
Solution:
1. In the fig.30.48 below:
• AF is the man standing at the foot of the tower
• GH is the man standing on the top of the tower
• GF is the tower and BE is the hill
2. In ⊿ADB, tan 60 = BD⁄AD ⟹ BD = AD × tan 60 ⟹ BD = FE × tan 60 = 40 × 1.7321 = 69.28 m
3. In ⊿GCB, tan 50 = BC⁄GC ⟹ BC = GC × tan 50 ⟹ BC = FE × tan 50 = 40 × 1.1917 = 47.67 m
4. Height of the tower = FH = FG - GH
5. FG = BE - BC
• BE = BD + DE = 69.28+1.75 = 71.03 m
• So FG = 71.03 - 47.67 = 23.36 m
6. Substituting this value of FG in (4) we get:
• Height of tower = FH = 23.36-1.75 = 21.61 m
7. Height of hill = BE = BD + DE = 69.28+1.75 = 71.03 m
Solved example 30.30
A 1.5 m tall boy saw the top of a building under construction at an elevation of 30o. The completed building was 10 m higher and the boy saw it's top at an elevation of 60o from the same spot. What is the height of the building?
Solution:
1. In the fig.30.49 below:
• AF is the boy
• CE is the building under construction
• BE is the building whose construction is completed
2. In ⊿ADB, tan 60 = BD⁄AD ⟹ AD = BD⁄tan 60
3. In ⊿ADC, tan 30 = CD⁄AD ⟹ AD = CD⁄tan 30
4. Equating (2) and (3) we get:
BD⁄tan 60 = CD⁄tan 30
5. But we have: CD = BD-10
Substituting this in (4) we get:
BD⁄tan 60 = (BD-10)⁄tan 30 ⟹ BD tan 30 = (BD-10) tan 60
⟹ BD tan 30 = BD tan 60 - 10 tan 60 ⟹ BD (tan 60 - tan 30) = 10 tan 60
⟹ BD (1.7320-0.5773) = 10 × 1.7320 ⟹ BD × 1.1547 = 17.32
⟹ BD = 14.999 = 15 m
6. So height of the completed building = BD + DE = 15+1.5 = 16.5 m
Solved example 30.31
When the sun is at an elevation of 40o, the length of the shadow of a tree is 18 m. What is the height of the tree?
Solution:
Consider fig.30.50(a) below:
• A sun ray is falling at point C.
• The ray makes an angle 40o with the horizontal.
♦ That is., at that particular time, the sun is at an elevation of 40o.
• The sun rays come from a very large distance. So all the sun rays are parallel.
■ Thus, at any particular time, the angle of elevation of the sun will be the same, whatever be the height of the observer.
• In the fig.a, AB is the ground level. A is a point on the ground. The angle of elevation for the sun will be 40o at A also. Based on thie above information we can solve this problem.
1. Consider fig.b. EF is the tree. The sun ray passing through the top point E will cast it's shadow at G. So GF is the shadow of the tree
2. We need the ⊿EFG only
tan 40 = EF⁄GF ⟹ EF = GF × tan 40 = 18 × 0.8391 = 15.10 m
Solved example 30.32
A man 1.8 m tall standing at the top of a telephone tower, saw the top of a 10 m high building at a depression of 40o and the base of the building at a depression of 60o. What is the height of the tower? How far is it from the building?
Solution:
• In the fig.30.51 below, AE is the man and ED is the telephone tower. BC is the building
1. In ⊿ACF, tan 60 = CF⁄AF ⟹ AF = CF⁄tan 60
2. In ⊿ABF, tan 40 = BF⁄AF ⟹ AF = BF⁄tan 40
3. Equating (1) and (2) we get:
CF⁄tan 60 = BF⁄tan 40
4. But we have: BF = CF-10
Substituting this in (3) we get:
CF⁄tan 60 = (CF-10)⁄tan 40 ⟹ CF tan 40 = (CF-10) tan 60
⟹ CF tan 40 = CF tan 60 - 10 tan 60 ⟹ CF (tan 60 - tan 40) = 10 tan 60
⟹ CF (1.7320-0.8391) = 10 × 1.7320 ⟹ CF × 0.8929 = 17.32
⟹ CF = 19.4 m
5. From (1) we get: AF = CF⁄tan 60 ⟹ AF = 19.4⁄1.7320 = 11.2 m
6. So distance between the tower and the building = CD = AF = 11.2 m
7. Height of the tower = DE = DA-EA
⟹ DE = CF-EA = 19.4 - 1.8 = 17.6 m
Solved example 30.33
From the top of an electric post, two wires are stretched to either sides and fixed to the ground, 25 m apart. The wires make angles 55o and 40o with the ground. What is the height of the post?
Solution:
• In the fig.30.52 below, CD is the post, CA and CB are the wires
1. In ⊿ADC, tan 55 = CD⁄AD ⟹ CD = AD × tan 55
2. In ⊿BDC, tan 40= CD⁄BD ⟹ CD = BD × tan 40
3. Equating (1) and (2) we get: AD tan 55 = BD tan 40
4. But BD = (25-AD)
5. Substituting this value of BD in (3) we get:
AD tan 55 = (25-AD) tan 40 ⟹ AD tan 55 = 25 tan 40 - AD tan 40
⟹ AD(tan 55 + tan 40) = 25 × tan 40
⟹ AD(1.4281+0.8391) = 25 × 0.8391 ⟹ AD × 2.2672 = 20.9775
⟹ AD = 9.2526 m
6. Substituting this value of AD in (1) we get:
CD = 9.2526 × tan 55 = 9.2526 × 1.4281 = 13.214 m
Solved example 30.34
When the sun is at an elevation of 35o, the shadow of a tree is 10 m long. What would be the length of the shadow, when the sun is at an elevation of 25o?
Solution:
• This is a problem similar to the solved example 30.31 that we saw above. See fig.30.53 below:
• AC is the sun ray when the elevation of the sun is 35
• CD is the sun ray when the elevation of the sun is 25
• CB is the tree
• We have to calculate BD
1. In ⊿ABC, tan 35 = BC⁄AB ⟹ BC = AB × tan 35 = 10 tan 35 = 10 × 0.7002 = 7.002 m
2. In ⊿DBC, tan 25= BC⁄BD ⟹ BD = BC⁄tan 25 = 7.002⁄0.4663 = 15.016 m
Solved example 30.29
A 1.75 metre tall man, standing at the foot of a tower, sees the top of a hill 40 metres away at an elevation of 60o. Climbing to the top of the tower, he sees it at an elevation of 50o. Calculate the heights of the tower and the hill.
Solution:
1. In the fig.30.48 below:
• AF is the man standing at the foot of the tower
• GH is the man standing on the top of the tower
• GF is the tower and BE is the hill
 |
| Fig.30.48 |
3. In ⊿GCB, tan 50 = BC⁄GC ⟹ BC = GC × tan 50 ⟹ BC = FE × tan 50 = 40 × 1.1917 = 47.67 m
4. Height of the tower = FH = FG - GH
5. FG = BE - BC
• BE = BD + DE = 69.28+1.75 = 71.03 m
• So FG = 71.03 - 47.67 = 23.36 m
6. Substituting this value of FG in (4) we get:
• Height of tower = FH = 23.36-1.75 = 21.61 m
7. Height of hill = BE = BD + DE = 69.28+1.75 = 71.03 m
Solved example 30.30
A 1.5 m tall boy saw the top of a building under construction at an elevation of 30o. The completed building was 10 m higher and the boy saw it's top at an elevation of 60o from the same spot. What is the height of the building?
Solution:
1. In the fig.30.49 below:
• AF is the boy
• CE is the building under construction
• BE is the building whose construction is completed
 |
| Fig.30.49 |
3. In ⊿ADC, tan 30 = CD⁄AD ⟹ AD = CD⁄tan 30
4. Equating (2) and (3) we get:
BD⁄tan 60 = CD⁄tan 30
5. But we have: CD = BD-10
Substituting this in (4) we get:
BD⁄tan 60 = (BD-10)⁄tan 30 ⟹ BD tan 30 = (BD-10) tan 60
⟹ BD tan 30 = BD tan 60 - 10 tan 60 ⟹ BD (tan 60 - tan 30) = 10 tan 60
⟹ BD (1.7320-0.5773) = 10 × 1.7320 ⟹ BD × 1.1547 = 17.32
⟹ BD = 14.999 = 15 m
6. So height of the completed building = BD + DE = 15+1.5 = 16.5 m
Solved example 30.31
When the sun is at an elevation of 40o, the length of the shadow of a tree is 18 m. What is the height of the tree?
Solution:
Consider fig.30.50(a) below:
 |
| Fig.30.50 |
• The ray makes an angle 40o with the horizontal.
♦ That is., at that particular time, the sun is at an elevation of 40o.
• The sun rays come from a very large distance. So all the sun rays are parallel.
■ Thus, at any particular time, the angle of elevation of the sun will be the same, whatever be the height of the observer.
• In the fig.a, AB is the ground level. A is a point on the ground. The angle of elevation for the sun will be 40o at A also. Based on thie above information we can solve this problem.
1. Consider fig.b. EF is the tree. The sun ray passing through the top point E will cast it's shadow at G. So GF is the shadow of the tree
2. We need the ⊿EFG only
tan 40 = EF⁄GF ⟹ EF = GF × tan 40 = 18 × 0.8391 = 15.10 m
Solved example 30.32
A man 1.8 m tall standing at the top of a telephone tower, saw the top of a 10 m high building at a depression of 40o and the base of the building at a depression of 60o. What is the height of the tower? How far is it from the building?
Solution:
• In the fig.30.51 below, AE is the man and ED is the telephone tower. BC is the building
 |
| Fig.30.51 |
2. In ⊿ABF, tan 40 = BF⁄AF ⟹ AF = BF⁄tan 40
3. Equating (1) and (2) we get:
CF⁄tan 60 = BF⁄tan 40
4. But we have: BF = CF-10
Substituting this in (3) we get:
CF⁄tan 60 = (CF-10)⁄tan 40 ⟹ CF tan 40 = (CF-10) tan 60
⟹ CF tan 40 = CF tan 60 - 10 tan 60 ⟹ CF (tan 60 - tan 40) = 10 tan 60
⟹ CF (1.7320-0.8391) = 10 × 1.7320 ⟹ CF × 0.8929 = 17.32
⟹ CF = 19.4 m
5. From (1) we get: AF = CF⁄tan 60 ⟹ AF = 19.4⁄1.7320 = 11.2 m
6. So distance between the tower and the building = CD = AF = 11.2 m
7. Height of the tower = DE = DA-EA
⟹ DE = CF-EA = 19.4 - 1.8 = 17.6 m
Solved example 30.33
From the top of an electric post, two wires are stretched to either sides and fixed to the ground, 25 m apart. The wires make angles 55o and 40o with the ground. What is the height of the post?
Solution:
• In the fig.30.52 below, CD is the post, CA and CB are the wires
 |
| Fig.30.52 |
2. In ⊿BDC, tan 40= CD⁄BD ⟹ CD = BD × tan 40
3. Equating (1) and (2) we get: AD tan 55 = BD tan 40
4. But BD = (25-AD)
5. Substituting this value of BD in (3) we get:
AD tan 55 = (25-AD) tan 40 ⟹ AD tan 55 = 25 tan 40 - AD tan 40
⟹ AD(tan 55 + tan 40) = 25 × tan 40
⟹ AD(1.4281+0.8391) = 25 × 0.8391 ⟹ AD × 2.2672 = 20.9775
⟹ AD = 9.2526 m
6. Substituting this value of AD in (1) we get:
CD = 9.2526 × tan 55 = 9.2526 × 1.4281 = 13.214 m
Solved example 30.34
When the sun is at an elevation of 35o, the shadow of a tree is 10 m long. What would be the length of the shadow, when the sun is at an elevation of 25o?
Solution:
• This is a problem similar to the solved example 30.31 that we saw above. See fig.30.53 below:
 |
| Fig.30.53 |
• CD is the sun ray when the elevation of the sun is 25
• CB is the tree
• We have to calculate BD
1. In ⊿ABC, tan 35 = BC⁄AB ⟹ BC = AB × tan 35 = 10 tan 35 = 10 × 0.7002 = 7.002 m
2. In ⊿DBC, tan 25= BC⁄BD ⟹ BD = BC⁄tan 25 = 7.002⁄0.4663 = 15.016 m
This completes our present discussion on Trigonometry. In the next section we will see Coordinates.
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