In the previous section we saw trigonometry related to circles. In the discussion so far in this chapter, we have seen problems related to sine and cosine. We have seen the basic details of tangent also. Details here. In this section, we will see some problems related to tangent.
An example:
• Some steps are built on the side of a hill. See fig.30.33(a) below:
• The steps lead up to the top of the hill. But all the steps are not shown in the fig. We need only the first few steps for our present problem.
• A cone (coloured in yellow) is placed on the third step. We want to know the level at which the base of the cone is situated. In other words, what is the value of 'h'?
• For solving this problem, we are given two information:
♦ The tread of each step is 20 cm
♦ The angle which the side of the hill makes with the horizontal ground surface is 35o
Solution:
1. Drop a perpendicular CB from the edge of the third step onto the ground surface. This is shown by the red dashed line in fig.(b)
• Now we have a right triangle: ⊿ABC.
2. Base AB will be equal to 3 times the tread. So we get AB = 3 × 20 = 60 cm
3. In this right triangle,
tan 35 = opposite side⁄adjacent side = BC⁄AB = h⁄60 ⟹ h = 60 × tan 35 ⟹ h = 60 × 0.7002 = 42.01 cm
How high is the top end of the iron rod from the ground?
Solution:
1. The wall, iron rod and the ground surface together will give a right triangle as shown in fig.(b)
We have tan 40 = opposite side⁄adjacent side = h⁄2 ⟹ h = 2 × tan 40 ⟹ h = 2 × 0.8391 = 1.68 m
Solved example 30.22
Three rectangles are cut along the diagonals as shown in the figs.30.35(a, b and c) below:
One large rectangle and two smaller rectangles. The two smaller rectangles are identical. The 6 pieces so obtained are rearranged to form a regular pentagon. See. fig.(d). If the sides of the pentagon are to be 30 cm, what should be the length and breadth of the rectangles
Solution:
1. First we want the interior angles of the pentagon. Since it is a regular pentagon, all sides will be equal. All angles will also be equal.
• We have: Sum of interior angles of any regular polygon = (n-2)×180.
Where n is the number of sides.
• For a regular pentagon, n = 5. So the sum of all the interior angles = (5-2)×180 = 3×180 = 540o
♦ So angle at each vertex = 540⁄5 = 108o
2. Now consider the yellow and magenta triangles in fig.30.36 below:
3. They are identical. So the edge AB will bisect the interior angle (at vertex A) of the pentagon. So we get: ∠CAB = ∠DAB = 108⁄2 = 54o
4. Consider ⊿ABC.
sin 54 = opposite side⁄hypotenuse = BC⁄AC = BC⁄30 ⟹ BC = 30 × sin 54
⟹ BC = 30 × 0.8090 = 24.27 cm
■ So we get: Length of the small rectangles = 24.27 cm
5. Again consider ⊿ABC.
cos 54 = adjacent side⁄hypotenuse = AB⁄AC = AB⁄30 ⟹ AB = 30 × cos 54
⟹ AB = 30 × 0.5878 = 17.63 cm
■ So we get: Width of the small rectangles = 17.63 cm
6. Now consider the vertex C of the pentagon. The yellow and green triangles are identical. So each of them will contribute 36o at the vertex C
7. The blue and red triangles are identical. So they will contribute equally at vertex C
So we can write:
2×36 + 2 × contribution from blue triangle = 108
⟹ 72 + 2 contribution from blue triangle = 108
⟹ 2 × contribution from blue triangle = 36
⟹ contribution from blue triangle = 36⁄2 = 18o
• So contribution from red triangle is also 18o
8. Now consider triangle DEC. It is right angled at E. So we get:
∠CDE + 90 + 18 = 180 ⟹ ∠CDE + 108 = 180 ⟹ ∠CDE = 72o
9. tan 72 = CE⁄DE
• But DE = half of the side of the pentagon = 15 cm
• So we get: tan 72 = CE⁄15 ⟹ CE = 15 × tan 72 ⟹ CE = 15 × 3.078 = 46.17 cm
■ So length of the large rectangle is 46.17 cm and it's width is 15 cm
Solved example 30.23
In the fig.30.37(a) below, the vertical lines are equally spaced.
Prove that their heights are in arithmetic sequence
Solution:
1. In fig.(b), the vertical lines are marked as AE, BF, CG etc.,
2. These vertical lines are drawn at an equal spacing of 's' cm
• Let the initial distance OA be 'a' cm
3. Consider the first triangle OAE. Since AE is vertical, triangle OAE is right angled at A
• So we get: tan 40 = AE⁄a ⟹ AE = a tan 40
4. Consider the second triangle OBF. Since BF is vertical, triangle OBF is right angled at B
• So we get: tan 40 = BF⁄(a+s) ⟹ BF = (a+s) tan 40
5. Consider the third triangle OCG. Since CG is vertical, triangle OCG is right angled at C
• So we get: tan 40 = CG⁄(a+s+s) ⟹ BF = (a+s+s) tan 40 ⟹ BF = (a+2s) tan 40
6. Consider the fourth triangle ODH. Since DH is vertical, triangle ODH is right angled at D
• So we get: tan 40 = DH⁄(a+s+s+s) ⟹ DH = (a+s+s+s) tan 40 ⟹ DH = (a+3s) tan 40
7. Continuing like this, if there are more vertical lines, we will get their heights as:
(a+4s) tan 40, (a+5s) tan 40 etc.,
8. Let us write the heights as a sequence:
a tan 40, (a+s) tan 40, (a+2s) tan 40, (a+3s) tan 40 . . . ,
9. In the above sequence, only 4 terms are written. We can write more terms if we want, just by increasing the coefficient of 's'.
• Now we have to prove that this sequence is an arithmetic sequence
10. Fourth term - third term = (a+3s) tan 40 - (a+2s) tan 40
= tan 40[(a+3s)-(a+2s)]
= tan 40[a+3s-a-2s]
= tan 40[s]
= s tan 40
11. Third term - second term = (a+2s) tan 40 - (a+s) tan 40
= tan 40[(a+2s)-(a+s)]
= tan 40[a+2s-a-s]
= tan 40[s]
= s tan 40
12. Second term - first term = (a+s) tan 40 - (a) tan 40
= tan 40[(a+s)-(a)]
= tan 40[a+s-a]
= tan 40[s]
= s tan 40
13. So we find that the difference the difference between any term and it's preceding term is a constant.
• So it is an arithmetic sequence
• The common difference is s tan40.
Solved example 30.24
One side of a triangle is 6 cm and the angles at it's ends are 40 and 65. Calculate it's area
Solution:
1. The given data is shown in fig.30.38(a) below:
2. Let us name the triangle as ABC. Drop a perpendicular CD from the vertex C on to the base AB
• Then area of ΔABC = 1⁄2 × AB × CD
• So we have to find CD
3. In ΔADC we have:
tan 40 = CD⁄AD ⟹ AD = CD⁄tan 40
4. In ΔBDC we have:
tan 65 = CD⁄BD ⟹ BD = CD⁄tan 65
5. But AB = AD + BD ⟹ 6 = [CD⁄tan 40 + CD⁄tan 65] ⟹ 6 = [CD⁄tan 40 + CD⁄tan 65]
⟹ 6 = [CD⁄0.8391 + CD⁄2.1446]
[(CD×2.1446)⁄(0.8391× 2.1446) + (CD×0.8391)⁄(0.8391×2.1446)] = 6
⟹ [(CD×2.1446)⁄(0.8391× 2.1446) + (CD×0.8391)⁄(0.8391×2.1446)] = 6
⟹ [(CD×2.1446) + (CD×0.8391)] = 6×0.8391×2.1446
⟹ (2.9837 × CD) = 10.7972
⟹ CD = 3.62 cm
6. So area = 1⁄2 × AB × CD = 1⁄2 × 6 × 3.62 = 10.86 cm2
Solved example 30.25
In the fig.30.39(a) below, prove that h = √(xy)
Solution:
1. The given triangle is named as ABC in fig.(b). Drop a perpendicular CD from the vertex C onto the base AB
2. ADC is a right triangle.
We have: ∠CAD + 90 + ∠ACD = 180 ⟹ 30 + 90 + ∠ACD = 180o ⟹ ∠ACD = 60o
3. So it is a 30, 60, 90 triangle
• CD is the shortest side (opposite to the smallest angle 30)
• AD is the medium side (opposite to the medium angle 60)
• AC is the largest side (opposite to the largest angle 90)
4. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• AD = √3 × CD ⟹ x = √3 × h = h√3 cm
5. BDC is a right triangle.
We have: ∠CBD + 90 + ∠BCD = 180 ⟹ 60 + 90 + ∠BCD = 180o ⟹ ∠BCD = 30o
6. So it is a 30, 60, 90 triangle
• BD is the shortest side (opposite to the smallest angle 30)
• CD is the medium side (opposite to the medium angle 60)
• BC is the largest side (opposite to the largest angle 90)
7. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• CD = √3 × BD ⟹ h = √3 × y = y√3 ⟹ y = h⁄(√3)
8. From (4) we have: x = h√3
From (7) we have: y = h⁄(√3)
9. So xy = h√3 × [h⁄(√3)] = h2 ⟹ h = √(xy)
An example:
• Some steps are built on the side of a hill. See fig.30.33(a) below:
 |
| Fig.30.33 |
• A cone (coloured in yellow) is placed on the third step. We want to know the level at which the base of the cone is situated. In other words, what is the value of 'h'?
• For solving this problem, we are given two information:
♦ The tread of each step is 20 cm
♦ The angle which the side of the hill makes with the horizontal ground surface is 35o
Solution:
1. Drop a perpendicular CB from the edge of the third step onto the ground surface. This is shown by the red dashed line in fig.(b)
• Now we have a right triangle: ⊿ABC.
2. Base AB will be equal to 3 times the tread. So we get AB = 3 × 20 = 60 cm
3. In this right triangle,
tan 35 = opposite side⁄adjacent side = BC⁄AB = h⁄60 ⟹ h = 60 × tan 35 ⟹ h = 60 × 0.7002 = 42.01 cm
Now we will see a few solved examples like this:
Solved example 30.21
An iron rod of unknown length, leans against a wall. The bottom end of the iron rod is 2 m away from the wall. Also it makes an angle of 40o with the floor. See fig.30.34(a) below: |
| Fig.30.34 |
Solution:
1. The wall, iron rod and the ground surface together will give a right triangle as shown in fig.(b)
We have tan 40 = opposite side⁄adjacent side = h⁄2 ⟹ h = 2 × tan 40 ⟹ h = 2 × 0.8391 = 1.68 m
Solved example 30.22
Three rectangles are cut along the diagonals as shown in the figs.30.35(a, b and c) below:
 |
| Fig.30.35 |
Solution:
1. First we want the interior angles of the pentagon. Since it is a regular pentagon, all sides will be equal. All angles will also be equal.
• We have: Sum of interior angles of any regular polygon = (n-2)×180.
Where n is the number of sides.
• For a regular pentagon, n = 5. So the sum of all the interior angles = (5-2)×180 = 3×180 = 540o
♦ So angle at each vertex = 540⁄5 = 108o
2. Now consider the yellow and magenta triangles in fig.30.36 below:
 |
| Fig.30.36 |
4. Consider ⊿ABC.
sin 54 = opposite side⁄hypotenuse = BC⁄AC = BC⁄30 ⟹ BC = 30 × sin 54
⟹ BC = 30 × 0.8090 = 24.27 cm
■ So we get: Length of the small rectangles = 24.27 cm
5. Again consider ⊿ABC.
cos 54 = adjacent side⁄hypotenuse = AB⁄AC = AB⁄30 ⟹ AB = 30 × cos 54
⟹ AB = 30 × 0.5878 = 17.63 cm
■ So we get: Width of the small rectangles = 17.63 cm
6. Now consider the vertex C of the pentagon. The yellow and green triangles are identical. So each of them will contribute 36o at the vertex C
7. The blue and red triangles are identical. So they will contribute equally at vertex C
So we can write:
2×36 + 2 × contribution from blue triangle = 108
⟹ 72 + 2 contribution from blue triangle = 108
⟹ 2 × contribution from blue triangle = 36
⟹ contribution from blue triangle = 36⁄2 = 18o
• So contribution from red triangle is also 18o
8. Now consider triangle DEC. It is right angled at E. So we get:
∠CDE + 90 + 18 = 180 ⟹ ∠CDE + 108 = 180 ⟹ ∠CDE = 72o
9. tan 72 = CE⁄DE
• But DE = half of the side of the pentagon = 15 cm
• So we get: tan 72 = CE⁄15 ⟹ CE = 15 × tan 72 ⟹ CE = 15 × 3.078 = 46.17 cm
■ So length of the large rectangle is 46.17 cm and it's width is 15 cm
Solved example 30.23
In the fig.30.37(a) below, the vertical lines are equally spaced.
 |
| Fig.30.37 |
Solution:
1. In fig.(b), the vertical lines are marked as AE, BF, CG etc.,
2. These vertical lines are drawn at an equal spacing of 's' cm
• Let the initial distance OA be 'a' cm
3. Consider the first triangle OAE. Since AE is vertical, triangle OAE is right angled at A
• So we get: tan 40 = AE⁄a ⟹ AE = a tan 40
4. Consider the second triangle OBF. Since BF is vertical, triangle OBF is right angled at B
• So we get: tan 40 = BF⁄(a+s) ⟹ BF = (a+s) tan 40
5. Consider the third triangle OCG. Since CG is vertical, triangle OCG is right angled at C
• So we get: tan 40 = CG⁄(a+s+s) ⟹ BF = (a+s+s) tan 40 ⟹ BF = (a+2s) tan 40
6. Consider the fourth triangle ODH. Since DH is vertical, triangle ODH is right angled at D
• So we get: tan 40 = DH⁄(a+s+s+s) ⟹ DH = (a+s+s+s) tan 40 ⟹ DH = (a+3s) tan 40
7. Continuing like this, if there are more vertical lines, we will get their heights as:
(a+4s) tan 40, (a+5s) tan 40 etc.,
8. Let us write the heights as a sequence:
a tan 40, (a+s) tan 40, (a+2s) tan 40, (a+3s) tan 40 . . . ,
9. In the above sequence, only 4 terms are written. We can write more terms if we want, just by increasing the coefficient of 's'.
• Now we have to prove that this sequence is an arithmetic sequence
10. Fourth term - third term = (a+3s) tan 40 - (a+2s) tan 40
= tan 40[(a+3s)-(a+2s)]
= tan 40[a+3s-a-2s]
= tan 40[s]
= s tan 40
11. Third term - second term = (a+2s) tan 40 - (a+s) tan 40
= tan 40[(a+2s)-(a+s)]
= tan 40[a+2s-a-s]
= tan 40[s]
= s tan 40
12. Second term - first term = (a+s) tan 40 - (a) tan 40
= tan 40[(a+s)-(a)]
= tan 40[a+s-a]
= tan 40[s]
= s tan 40
13. So we find that the difference the difference between any term and it's preceding term is a constant.
• So it is an arithmetic sequence
• The common difference is s tan40.
Solved example 30.24
One side of a triangle is 6 cm and the angles at it's ends are 40 and 65. Calculate it's area
Solution:
1. The given data is shown in fig.30.38(a) below:
 |
| Fig.30.38 |
• Then area of ΔABC = 1⁄2 × AB × CD
• So we have to find CD
3. In ΔADC we have:
tan 40 = CD⁄AD ⟹ AD = CD⁄tan 40
4. In ΔBDC we have:
tan 65 = CD⁄BD ⟹ BD = CD⁄tan 65
5. But AB = AD + BD ⟹ 6 = [CD⁄tan 40 + CD⁄tan 65] ⟹ 6 = [CD⁄tan 40 + CD⁄tan 65]
⟹ 6 = [CD⁄0.8391 + CD⁄2.1446]
[(CD×2.1446)⁄(0.8391× 2.1446) + (CD×0.8391)⁄(0.8391×2.1446)] = 6
⟹ [(CD×2.1446)⁄(0.8391× 2.1446) + (CD×0.8391)⁄(0.8391×2.1446)] = 6
⟹ [(CD×2.1446) + (CD×0.8391)] = 6×0.8391×2.1446
⟹ (2.9837 × CD) = 10.7972
⟹ CD = 3.62 cm
6. So area = 1⁄2 × AB × CD = 1⁄2 × 6 × 3.62 = 10.86 cm2
Solved example 30.25
In the fig.30.39(a) below, prove that h = √(xy)
 |
| Fig.30.39 |
1. The given triangle is named as ABC in fig.(b). Drop a perpendicular CD from the vertex C onto the base AB
2. ADC is a right triangle.
We have: ∠CAD + 90 + ∠ACD = 180 ⟹ 30 + 90 + ∠ACD = 180o ⟹ ∠ACD = 60o
3. So it is a 30, 60, 90 triangle
• CD is the shortest side (opposite to the smallest angle 30)
• AD is the medium side (opposite to the medium angle 60)
• AC is the largest side (opposite to the largest angle 90)
4. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• AD = √3 × CD ⟹ x = √3 × h = h√3 cm
5. BDC is a right triangle.
We have: ∠CBD + 90 + ∠BCD = 180 ⟹ 60 + 90 + ∠BCD = 180o ⟹ ∠BCD = 30o
6. So it is a 30, 60, 90 triangle
• BD is the shortest side (opposite to the smallest angle 30)
• CD is the medium side (opposite to the medium angle 60)
• BC is the largest side (opposite to the largest angle 90)
7. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• CD = √3 × BD ⟹ h = √3 × y = y√3 ⟹ y = h⁄(√3)
8. From (4) we have: x = h√3
From (7) we have: y = h⁄(√3)
9. So xy = h√3 × [h⁄(√3)] = h2 ⟹ h = √(xy)
In the next section we will see another application of tangent.
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