In the previous section we completed a discussion on cosine of an acute angle. In this section we will see tangent. For that, let us do a fifth experiment.
Experiment 5:
1. A person goes to the ⊿ABC in fig.30.10(a) below. He takes up position at vertex A
[Note that, fig.30.10 below is the same fig.30.6 that we saw at the beginning of the previous section, for Experiment 3]
2. At the vertex A, the angle is 37o
• For this 37o, the opposite side is BC
• For this 37o, the adjacent side is AB
• Side AC is also an adjacent side of 37o. But since AC is the hypotenuse, we will call it 'hypotenuse' itself. Because, which ever is the vertex that we consider, hypotenuse of a triangle will not change. But 'adjacent' and 'opposite' will change with respect to the vertex under consideration.
3. So we have:
• Opposite side of 37o = BC = 4.0874 m
• Adjacent side of 37o = AB = 5.4242 m
• Hypotenuse of the triangle = AC = 6.7918 m
4. Let us take the ratio: opposite side of 37⁄adjacent side of 37
[Note that, in all the previous experiments, we used 'hypotenuse' in the denominator]
• We get: BC⁄AB = 4.0874⁄5.4242 = 0.7535
• Note down this value. We will continue our experiment:
5. The person now goes to ⊿MNO in fig.30.10(b). He takes up position at vertex M. He takes up this particular position because, the angle there is 37o, the same value in the previous ⊿ABC
6. At the vertex M, the angle is 37o
• For this 37o, the opposite side is ON
• For this 37o, the adjacent side is MN
• Side OM is also an adjacent side. But since OM is the hypotenuse, we will call it 'hypotenuse' itself. 7. So we have:
• Opposite side of 37o = ON = 5.8258 m
• Adjacent side of 37o = MN = 7.7311 m
• Hypotenuse of the triangle = OM = 9.6804 m
8. Let us take the ratio: opposite side of 37⁄adjacent side of 37
• We get: ON⁄MN = 5.8258⁄7.7311 = 0.7535
■ This is the same value that we obtained in (4)
• Note down this value. We will continue our experiment:
9. The person now goes to ⊿PQR in fig.30.10(c). He takes up position at vertex P. He takes up this particular position because, the angle there is 37o, the same value in the previous triangles
10. At the vertex P, the angle is 37o
• For this 37o, the opposite side is RQ
• For this 37o, the adjacent side is PR
• Side PQ is also an adjacent side. But since PQ is the hypotenuse, we will call it 'hypotenuse' itself.
11. So we have:
• Opposite side of 37o = RQ = 5.2661 m
• Adjacent side of 37o = PR = 6.9884 m
• Hypotenuse of the triangle = PQ = 8.7504 m
12. Let us take the ratio: opposite side of 37⁄adjacent side of 37
• We get: RQ⁄PR = 5.2661⁄6.9884 = 0.7535
■ This is the same value that we obtained in (4) & (8)
• Note down this value. We will continue our experiment:
13. The person now goes to ⊿UVW in fig.30.10(d). He takes up position at vertex V. He takes up this particular position because, the angle there is 37o, the same value in the previous triangles
14. At the vertex V, the angle is 37o
• For this 37o, the opposite side is UW
• For this 37o, the adjacent side is VW
• Side UV is also an adjacent side. But since UV is the hypotenuse, we will call it 'hypotenuse' itself.
15. So we have:
• Opposite side of 37o = UW = 3.7913 m
• Adjacent side of 37o = VW = 5.0312 m
• Hypotenuse of the triangle = UV = 6.2998 m
16. Let us take the ratio: opposite side of 37⁄adjacent side of 37
• We get: UW⁄VW = 3.7913⁄5.0312 = 0.7535
■ This is the same value that we obtained in (4), (8) & (12)
• Experiment 3 is now complete
• All of them had one acute angle of 37o
• In all of them we took the ratio: opposite side of 37⁄adjacent side of 37
• In all of them we got the same value 0.7935
■ So we can conclude:
• Whenever we get a right triangle with one acute angle 37o, we can write:
opposite side of 37 in that triangle⁄adjacent side of 37 in that triangle = 0.7935
• We don't even have to do any calculations with the sides
• That means, opposite side of 37⁄adjacent side of 37 is a constant 0.7935
• Later in this section, we will see the reason for this 'constant nature'. In fact, corresponding to 37o, the value 0.7935 is written in standard tables published by the scientific community. We will see such tables also later.
1. A person goes to the ⊿ABC in fig.30.11(a) below. He takes up position at vertex A
2. At the vertex A, the angle is 68o
• For this 68o, the opposite side is BC
• For this 68o, the adjacent side is AB
• Side AC is also an adjacent side of 68o. But since AC is the hypotenuse, we will call it 'hypotenuse' itself. Because, which ever is the vertex that we consider, hypotenuse of a triangle will not change. But 'adjacent' and 'opposite' will change with respect to the vertex under consideration.
3. So we have:
• Opposite side of 68o = BC = 10.2909 m
• Adjacent side of 68o = AB = 4.1578 m
• Hypotenuse of the triangle = AC = 11.0991 m
4. Let us take the ratio: opposite side of 68⁄adjacent of 68
• We get: BC⁄AB = 10.2909⁄4.1578 = 2.4751
• Note down this value. We will continue our experiment:
5. The person now goes to ⊿MNO in fig.30.7(b). He takes up position at vertex M. He takes up this particular position because, the angle there is 68o, the same value in the previous ⊿ABC
6. At the vertex M, the angle is 68o
• For this 68o, the opposite side is ON
• For this 68o, the adjacent side is MN
• Side OM is also an adjacent side. But since OM is the hypotenuse, we will call it 'hypotenuse' itself.
7. So we have:
• Opposite side of 68o = ON = 5.1201 m
• Adjacent side of 68o = MN = 2.0687 m
• Hypotenuse of the triangle = OM = 5.5222 m
8. Let us take the ratio: opposite side of 68⁄adjacent of 68
• We get: ON⁄MN = 5.1201⁄2.0687 = 2.4751
■ This is the same value that we obtained in (4)
• Note down this value. We will continue our experiment:
9. The person now goes to ⊿PQR in fig.30.7(c). He takes up position at vertex P. He takes up this particular position because, the angle there is 68o, the same value in the previous triangles
10. At the vertex P, the angle is 68o
• For this 68o, the opposite side is RQ
• For this 68o, the adjacent side is PR
• Side PQ is also an adjacent side. But since PQ is the hypotenuse, we will call it 'hypotenuse' itself.
11. So we have:
• Opposite side of 68o = RQ = 8.1494 m
• Adjacent side of 68o = PR = 3.2926 m
• Hypotenuse of the triangle = PQ = 8.7894 m
12. Let us take the ratio: opposite side of 68⁄adjacent of 68
• We get: RQ⁄PR = 8.1494⁄3.2926 = 2.4751
■ This is the same value that we obtained in (4) & (8)
• Note down this value. We will continue our experiment:
13. The person now goes to ⊿UVW in fig.30.7(d). He takes up position at vertex V. He takes up this particular position because, the angle there is 68o, the same value in the previous triangles
14. At the vertex V, the angle is 68o
• For this 68o, the opposite side is VW
• For this 68o, the adjacent side is UW
• Side UV is also an adjacent side. But since UV is the hypotenuse, we will call it 'hypotenuse' itself.
15. So we have:
• Opposite side of 68o = VW = 15.4954 m
• Adjacent side of 68o = UW = 6.2606 m
• Hypotenuse of the triangle = UV = 16.7123 m
16. Let us take the ratio: opposite side of 68⁄adjacent of 68
• We get: VW⁄UW = 15.4954⁄6.2606 = 2.4751
■ This is the same value that we obtained in (4), (8) & (12)
• Experiment 4 is now complete
• All of them had one acute angle of 68o
• In all of them we took the ratio: opposite side of 68⁄adjacent of 68
• In all of them we got the same value 2.4751
■ So we can conclude:
• Whenever we get a right triangle with one acute angle 68o, we can write:
opposite side of 68 in that triangle⁄adjacent of 68 in that triangle = 2.4751
• We don't even have to do any calculations with the sides
• That means, opposite side of 68⁄adjacent of 68 is a constant 2.4751
• Later in this section, we will see the reason for this 'constant nature'. In fact, corresponding to 68o, the value 2.4751 is written in standard tables published by the scientific community. We will see such tables also later.
Let us write a summary of what we saw so far in this section. The summary can be written in just two lines:
1. We saw two acute angles. The first one 37o and the second one 68o.
2. For each of them, we saw that, there is a unique value for the ratio: opposite side⁄adjacent side
Now we will continue the discussion:
• These angles 37o and 68o were chosen at random. Whichever acute angle we take, we will get a similar result.
• That is., Take any acute angle. Let us represent it using 'θ' (Greek letter 'theta'). 'θ' can take any acute angle value like 2o, 5o, 35o, 72o, 1o, 25.2o, 15.4o . . .
• Millions of trillions of right triangles are possible with that particular acute angle θ.
• In all of those right triangles, the ratio:
opposite side of θ in that triangle⁄adjacent side of θ in that triangle will be same.
Let us see the reason for this:
1. Consider any acute angle θ. Draw any two possible right triangles with one angle θ. They are shown in the fig.30.12 below:
If one angle is θ, the other angle will be (90-θ). So the two triangles ⊿ABC and ⊿PQR are similar
= side opposite (90-θ) in ΔABC⁄side opposite (90-θ) in ΔPQR
= side opposite 90 in ΔABC⁄side opposite 90 in ΔPQR
5. So we get: BC⁄QR = AB⁄PR = AC⁄PQ
• opposite side of 1o⁄adjacent side of 1o = a unique value for 1o.
In the standard tables it is given as 0.0175
• opposite side of 2o⁄adjacent side of 2o = a unique value for 2o.
In the standard tables it is given as 0.0349
• opposite side of 3o⁄adjacent side of 3o = a unique value for 3o.
In the standard tables it is given as 0.0524
so on...
■ In general, for any acute angle θ, whenever we want the ratio opposite side of θ⁄adjacent side of θ , we can straight away look it up in the table, and write down the value.
Because of this 'constant nature' of the ratio, early mathematicians decided to give a special name. That special name is: 'Tangent'.
• opposite side of θ⁄adjacent side of θ is denoted as: 'tangent of θ'. It is abbreviated as 'tan θ'.
• So, from experiment 5, we have: tan 37 = 0.7986
• from experiment 6, we have: tan 68 = 2.4751
• A standard table that gives the tangent value of all acute angles 1, 2, 3, . . can be seen here.
Let us now see a very interesting result:
1. We have seen that sin θ = opposite side⁄hypotenuse
2. We have seen that cos θ = adjacent side⁄hypotenuse
3. Let us take the ratio of the above two. That is.,
sin θ⁄cos θ = opposite side⁄hypotenuse ÷ adjacent side ⁄hypotenuse
= opposite side⁄hypotenuse × hypotenuse ⁄adjacent side
= opposite side⁄adjacent side
4. But is opposite side⁄adjacent side tan θ. So we can write:
tan θ = sin θ⁄cos θ
5. sin θ is a constant. Let it be k1. cos θ is also a constant. Let it be k2
Then tan = sin θ⁄cos θ = k1⁄k2
6. But k1⁄k2 will be a third constant. Let it be k3. So we get:
tan θ = a constant k3
7. So we can use this method to prove that tan θ is a constant. Thus avoiding the long method that we saw above based on fig.30.12
Let us now see some solved examples:
Solved example 30.5
Given that tan 32 = 0.6249. Find the side AB of the ⊿ABC shown in fig.30.13(a) below:
Solution:
1. Given that tan 32 = 0.6249
2. So in any right triangle with one angle 32, the ratio
opposite side of that 32⁄adjacent side of that 32 = 0.6249
3. So, in our present triangle, we can write:
BC⁄AB = 0.6249 ⟹ 4⁄AB = 0.6249 ⟹ AB = 4⁄0.6249 = 6.4010 m
Solved example 30.6
Using the details given for ⊿PQR in fig.30.9(b), find the side UW of ⊿UVW. Assume that tables are not available for finding tangent values.
Solution:
1. First we calculate tan 56 from ⊿PQR. We get:
tan 56 = QR⁄PQ = 2⁄1.3490
2. In ⊿UVW, we can write:
tan 56 = opposite side of 56⁄adjacent side of 56 = UV⁄UW = 2.5⁄UW
4. tangent of an acute angle will be the same in any right triangle. So from (1) and (2) we get:
2⁄1.3490 = 2.5⁄UW = ⟹ UW = (2.5 × 1.3490)⁄2 = 1.6863
Experiment 5:
1. A person goes to the ⊿ABC in fig.30.10(a) below. He takes up position at vertex A
[Note that, fig.30.10 below is the same fig.30.6 that we saw at the beginning of the previous section, for Experiment 3]
 |
| Fig.30.10 |
• For this 37o, the opposite side is BC
• For this 37o, the adjacent side is AB
• Side AC is also an adjacent side of 37o. But since AC is the hypotenuse, we will call it 'hypotenuse' itself. Because, which ever is the vertex that we consider, hypotenuse of a triangle will not change. But 'adjacent' and 'opposite' will change with respect to the vertex under consideration.
3. So we have:
• Opposite side of 37o = BC = 4.0874 m
• Adjacent side of 37o = AB = 5.4242 m
• Hypotenuse of the triangle = AC = 6.7918 m
4. Let us take the ratio: opposite side of 37⁄adjacent side of 37
[Note that, in all the previous experiments, we used 'hypotenuse' in the denominator]
• We get: BC⁄AB = 4.0874⁄5.4242 = 0.7535
• Note down this value. We will continue our experiment:
5. The person now goes to ⊿MNO in fig.30.10(b). He takes up position at vertex M. He takes up this particular position because, the angle there is 37o, the same value in the previous ⊿ABC
6. At the vertex M, the angle is 37o
• For this 37o, the opposite side is ON
• For this 37o, the adjacent side is MN
• Side OM is also an adjacent side. But since OM is the hypotenuse, we will call it 'hypotenuse' itself. 7. So we have:
• Opposite side of 37o = ON = 5.8258 m
• Adjacent side of 37o = MN = 7.7311 m
• Hypotenuse of the triangle = OM = 9.6804 m
8. Let us take the ratio: opposite side of 37⁄adjacent side of 37
• We get: ON⁄MN = 5.8258⁄7.7311 = 0.7535
■ This is the same value that we obtained in (4)
• Note down this value. We will continue our experiment:
9. The person now goes to ⊿PQR in fig.30.10(c). He takes up position at vertex P. He takes up this particular position because, the angle there is 37o, the same value in the previous triangles
10. At the vertex P, the angle is 37o
• For this 37o, the opposite side is RQ
• For this 37o, the adjacent side is PR
• Side PQ is also an adjacent side. But since PQ is the hypotenuse, we will call it 'hypotenuse' itself.
11. So we have:
• Opposite side of 37o = RQ = 5.2661 m
• Adjacent side of 37o = PR = 6.9884 m
• Hypotenuse of the triangle = PQ = 8.7504 m
12. Let us take the ratio: opposite side of 37⁄adjacent side of 37
• We get: RQ⁄PR = 5.2661⁄6.9884 = 0.7535
■ This is the same value that we obtained in (4) & (8)
• Note down this value. We will continue our experiment:
13. The person now goes to ⊿UVW in fig.30.10(d). He takes up position at vertex V. He takes up this particular position because, the angle there is 37o, the same value in the previous triangles
14. At the vertex V, the angle is 37o
• For this 37o, the opposite side is UW
• For this 37o, the adjacent side is VW
• Side UV is also an adjacent side. But since UV is the hypotenuse, we will call it 'hypotenuse' itself.
15. So we have:
• Opposite side of 37o = UW = 3.7913 m
• Adjacent side of 37o = VW = 5.0312 m
• Hypotenuse of the triangle = UV = 6.2998 m
16. Let us take the ratio: opposite side of 37⁄adjacent side of 37
• We get: UW⁄VW = 3.7913⁄5.0312 = 0.7535
■ This is the same value that we obtained in (4), (8) & (12)
• Experiment 3 is now complete
We have done four trials. In all trials we got the same value. We will stop doing further trials and write a summary:
• We saw four right triangles• All of them had one acute angle of 37o
• In all of them we took the ratio: opposite side of 37⁄adjacent side of 37
• In all of them we got the same value 0.7935
■ So we can conclude:
• Whenever we get a right triangle with one acute angle 37o, we can write:
opposite side of 37 in that triangle⁄adjacent side of 37 in that triangle = 0.7935
• We don't even have to do any calculations with the sides
• That means, opposite side of 37⁄adjacent side of 37 is a constant 0.7935
• Later in this section, we will see the reason for this 'constant nature'. In fact, corresponding to 37o, the value 0.7935 is written in standard tables published by the scientific community. We will see such tables also later.
So we found the value for 37o. Let us try another angle:
Experiment 6:1. A person goes to the ⊿ABC in fig.30.11(a) below. He takes up position at vertex A
 |
| Fig.30.11 |
• For this 68o, the opposite side is BC
• For this 68o, the adjacent side is AB
• Side AC is also an adjacent side of 68o. But since AC is the hypotenuse, we will call it 'hypotenuse' itself. Because, which ever is the vertex that we consider, hypotenuse of a triangle will not change. But 'adjacent' and 'opposite' will change with respect to the vertex under consideration.
3. So we have:
• Opposite side of 68o = BC = 10.2909 m
• Adjacent side of 68o = AB = 4.1578 m
• Hypotenuse of the triangle = AC = 11.0991 m
4. Let us take the ratio: opposite side of 68⁄adjacent of 68
• We get: BC⁄AB = 10.2909⁄4.1578 = 2.4751
• Note down this value. We will continue our experiment:
5. The person now goes to ⊿MNO in fig.30.7(b). He takes up position at vertex M. He takes up this particular position because, the angle there is 68o, the same value in the previous ⊿ABC
6. At the vertex M, the angle is 68o
• For this 68o, the opposite side is ON
• For this 68o, the adjacent side is MN
• Side OM is also an adjacent side. But since OM is the hypotenuse, we will call it 'hypotenuse' itself.
7. So we have:
• Opposite side of 68o = ON = 5.1201 m
• Adjacent side of 68o = MN = 2.0687 m
• Hypotenuse of the triangle = OM = 5.5222 m
8. Let us take the ratio: opposite side of 68⁄adjacent of 68
• We get: ON⁄MN = 5.1201⁄2.0687 = 2.4751
■ This is the same value that we obtained in (4)
• Note down this value. We will continue our experiment:
9. The person now goes to ⊿PQR in fig.30.7(c). He takes up position at vertex P. He takes up this particular position because, the angle there is 68o, the same value in the previous triangles
10. At the vertex P, the angle is 68o
• For this 68o, the opposite side is RQ
• For this 68o, the adjacent side is PR
• Side PQ is also an adjacent side. But since PQ is the hypotenuse, we will call it 'hypotenuse' itself.
11. So we have:
• Opposite side of 68o = RQ = 8.1494 m
• Adjacent side of 68o = PR = 3.2926 m
• Hypotenuse of the triangle = PQ = 8.7894 m
12. Let us take the ratio: opposite side of 68⁄adjacent of 68
• We get: RQ⁄PR = 8.1494⁄3.2926 = 2.4751
■ This is the same value that we obtained in (4) & (8)
• Note down this value. We will continue our experiment:
13. The person now goes to ⊿UVW in fig.30.7(d). He takes up position at vertex V. He takes up this particular position because, the angle there is 68o, the same value in the previous triangles
14. At the vertex V, the angle is 68o
• For this 68o, the opposite side is VW
• For this 68o, the adjacent side is UW
• Side UV is also an adjacent side. But since UV is the hypotenuse, we will call it 'hypotenuse' itself.
15. So we have:
• Opposite side of 68o = VW = 15.4954 m
• Adjacent side of 68o = UW = 6.2606 m
• Hypotenuse of the triangle = UV = 16.7123 m
16. Let us take the ratio: opposite side of 68⁄adjacent of 68
• We get: VW⁄UW = 15.4954⁄6.2606 = 2.4751
■ This is the same value that we obtained in (4), (8) & (12)
• Experiment 4 is now complete
We have done four trials. In all trials we got the same value. We will stop doing further trials and write a summary:
• We saw four right triangles• All of them had one acute angle of 68o
• In all of them we took the ratio: opposite side of 68⁄adjacent of 68
• In all of them we got the same value 2.4751
■ So we can conclude:
• Whenever we get a right triangle with one acute angle 68o, we can write:
opposite side of 68 in that triangle⁄adjacent of 68 in that triangle = 2.4751
• We don't even have to do any calculations with the sides
• That means, opposite side of 68⁄adjacent of 68 is a constant 2.4751
• Later in this section, we will see the reason for this 'constant nature'. In fact, corresponding to 68o, the value 2.4751 is written in standard tables published by the scientific community. We will see such tables also later.
Let us write a summary of what we saw so far in this section. The summary can be written in just two lines:
1. We saw two acute angles. The first one 37o and the second one 68o.
2. For each of them, we saw that, there is a unique value for the ratio: opposite side⁄adjacent side
Now we will continue the discussion:
• These angles 37o and 68o were chosen at random. Whichever acute angle we take, we will get a similar result.
• That is., Take any acute angle. Let us represent it using 'θ' (Greek letter 'theta'). 'θ' can take any acute angle value like 2o, 5o, 35o, 72o, 1o, 25.2o, 15.4o . . .
• Millions of trillions of right triangles are possible with that particular acute angle θ.
• In all of those right triangles, the ratio:
opposite side of θ in that triangle⁄adjacent side of θ in that triangle will be same.
Let us see the reason for this:
1. Consider any acute angle θ. Draw any two possible right triangles with one angle θ. They are shown in the fig.30.12 below:
 |
| Fig.30.12 |
2. Now we apply a special property that is applicable to any two similar triangles (Details here):
side opposite smallest angle in ΔABC⁄side opposite smallest angle in ΔPQR
= side opposite medium angle in ΔABC⁄side opposite medium angle in ΔPQR
= side opposite largest angle in ΔABC⁄side opposite largest angle in ΔPQR
3. But angles in the two triangles are the same. That is.,
• Smallest angle in ΔABC = Smallest angle in ΔPQR
• Medium angle in ΔABC = Medium angle in ΔPQR
• Largest angle in ΔABC = Largest angle in ΔPQR
4. So we can write this:
• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,
side opposite θ in ΔABC⁄side opposite θ in ΔPQR = side opposite (90-θ) in ΔABC⁄side opposite (90-θ) in ΔPQR
= side opposite 90 in ΔABC⁄side opposite 90 in ΔPQR
5. So we get: BC⁄QR = AB⁄PR = AC⁄PQ
6. Take the first and the second ratios [Note that, in the discussion on sine, we took the first and the last ratios. And in the discussion on cosine, we took the second and the last ratios]. We get:
• BC⁄QR = AB⁄PR This can be written as:
• AB⁄AB = QR ⁄PR
• That is: [opposite side of θ⁄adjacent side of θ in ⊿ABC] = [opposite side of θ⁄adjacent side of θ in ⊿PQR] • Thus we find that the ratio is same in the two triangles.
7. So it is clear that the ratio opposite side of θ⁄adjacent side of θ will be a unique value. We can write:• BC⁄QR = AB⁄PR This can be written as:
• AB⁄AB = QR ⁄PR
• That is: [opposite side of θ⁄adjacent side of θ in ⊿ABC] = [opposite side of θ⁄adjacent side of θ in ⊿PQR] • Thus we find that the ratio is same in the two triangles.
• opposite side of 1o⁄adjacent side of 1o = a unique value for 1o.
In the standard tables it is given as 0.0175
• opposite side of 2o⁄adjacent side of 2o = a unique value for 2o.
In the standard tables it is given as 0.0349
• opposite side of 3o⁄adjacent side of 3o = a unique value for 3o.
In the standard tables it is given as 0.0524
so on...
■ In general, for any acute angle θ, whenever we want the ratio opposite side of θ⁄adjacent side of θ , we can straight away look it up in the table, and write down the value.
Because of this 'constant nature' of the ratio, early mathematicians decided to give a special name. That special name is: 'Tangent'.
• opposite side of θ⁄adjacent side of θ is denoted as: 'tangent of θ'. It is abbreviated as 'tan θ'.
• So, from experiment 5, we have: tan 37 = 0.7986
• from experiment 6, we have: tan 68 = 2.4751
• A standard table that gives the tangent value of all acute angles 1, 2, 3, . . can be seen here.
Let us now see a very interesting result:
1. We have seen that sin θ = opposite side⁄hypotenuse
2. We have seen that cos θ = adjacent side⁄hypotenuse
3. Let us take the ratio of the above two. That is.,
sin θ⁄cos θ = opposite side⁄hypotenuse ÷ adjacent side ⁄hypotenuse
= opposite side⁄hypotenuse × hypotenuse ⁄adjacent side
= opposite side⁄adjacent side
4. But is opposite side⁄adjacent side tan θ. So we can write:
tan θ = sin θ⁄cos θ
5. sin θ is a constant. Let it be k1. cos θ is also a constant. Let it be k2
Then tan = sin θ⁄cos θ = k1⁄k2
6. But k1⁄k2 will be a third constant. Let it be k3. So we get:
tan θ = a constant k3
7. So we can use this method to prove that tan θ is a constant. Thus avoiding the long method that we saw above based on fig.30.12
Let us now see some solved examples:
Solved example 30.5
Given that tan 32 = 0.6249. Find the side AB of the ⊿ABC shown in fig.30.13(a) below:
 |
| Fig.30.9 |
1. Given that tan 32 = 0.6249
2. So in any right triangle with one angle 32, the ratio
opposite side of that 32⁄adjacent side of that 32 = 0.6249
3. So, in our present triangle, we can write:
BC⁄AB = 0.6249 ⟹ 4⁄AB = 0.6249 ⟹ AB = 4⁄0.6249 = 6.4010 m
Solved example 30.6
Using the details given for ⊿PQR in fig.30.9(b), find the side UW of ⊿UVW. Assume that tables are not available for finding tangent values.
Solution:
1. First we calculate tan 56 from ⊿PQR. We get:
tan 56 = QR⁄PQ = 2⁄1.3490
2. In ⊿UVW, we can write:
tan 56 = opposite side of 56⁄adjacent side of 56 = UV⁄UW = 2.5⁄UW
4. tangent of an acute angle will be the same in any right triangle. So from (1) and (2) we get:
2⁄1.3490 = 2.5⁄UW = ⟹ UW = (2.5 × 1.3490)⁄2 = 1.6863
So we have completed the discussion on tangent. In the next section we will see some solved examples based on our discussions so far in this chapter.
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