In the previous section we completed a discussion on sine of an acute angle. In this section we will see cosine. For that, let us do a third experiment.
Experiment 3:
1. A person goes to the ⊿ABC in fig.30.6(a) below. He takes up position at vertex A
[Note that, fig.30.6 below is the same fig.30.2 that we saw at the beginning of the previous section, for Experiment 1]
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: adjacent side of 37⁄hypotenuse of the triangle
[Note that, in experiment 1, we used 'opposite side of 37' in the numerator]
• We get: AB⁄AC = 5.4242⁄6.7918 = 0.7986
• Note down this value. We will continue our experiment:
5. The person now goes to ⊿MNO in fig.30.6(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: adjacent side of 37⁄hypotenuse of the triangle
• We get: MN⁄OM = 7.7311⁄9.6804 = 0.7986
■ 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.30.6(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: adjacent side of 37⁄hypotenuse of the triangle
• We get: PR⁄PQ = 6.9884⁄8.7504 = 0.7986
■ 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.30.6(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: adjacent side of 37⁄hypotenuse of the triangle
• We get: VW⁄UV = 5.0312⁄6.2998 = 0.7986
■ 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: adjacent side of 37⁄hypotenuse of the triangle
• In all of them we got the same value 0.7986
■ So we can conclude:
• Whenever we get a right triangle with one acute angle 37o, we can write:
adjacent side of that 37⁄hypotenuse of that triangle = 0.7986
• We don't even have to do any calculations with the sides
• That means, adjacent side of 37⁄hypotenuse of the triangle is a constant 0.7986
• Later in this chapter, we will see the reason for this 'constant nature'. In fact, corresponding to 37o, the value 0.7986 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.30.7(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: adjacent side of 68⁄hypotenuse of the triangle
• We get: AB⁄AC = 4.1578⁄11.0991 = 0.3746
• 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: adjacent side of 68⁄hypotenuse of the triangle
• We get: MN⁄OM = 2.0687⁄5.5222 = 0.3746
■ 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: adjacent side of 68⁄hypotenuse of the triangle
• We get: PR⁄PQ = 3.2926⁄8.7894 = 0.3746
■ 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: adjacent side of 68⁄hypotenuse of the triangle
• We get: UW⁄UV = 6.2606⁄16.7123 = 0.3746
■ 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: adjacent side of 68⁄hypotenuse of the triangle
• In all of them we got the same value 0.3746
■ So we can conclude:
• Whenever we get a right triangle with one acute angle 68o, we can write:
adjacent side of that 68⁄hypotenuse of that triangle = 0.3746
• We don't even have to do any calculations with the sides
• That means, adjacent side of 68⁄hypotenuse of the triangle is a constant 0.3746
• Later in this chapter, we will see the reason for this 'constant nature'. In fact, corresponding to 68o, the value 0.3746 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. 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: adjacent side⁄hypotenuse
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: adjacent side of θ⁄hypotenuse of the 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.30.8 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
• adjacent side of 1o⁄hypotenuse = a unique value for 1o. In the standard tables it is given as 0.9998
• adjacent side of 2o⁄hypotenuse = a unique value for 2o. In the standard tables it is given as 0.9994
• adjacent side of 3o⁄hypotenuse = a unique value for 3o. In the standard tables it is given as 0.9986
so on...
■ In general, for any acute angle θ, whenever we want the ratio adjacent side of θ⁄hypotenuse , 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: 'Cosine'.
• adjacent side of θ⁄hypotenuse is denoted as: 'cosine of θ'. It is abbreviated as 'cos θ'.
• So, from experiment 3, we have: cos 37 = 0.7986
• from experiment 4, we have: cos 68 = 0.3746
• A standard table that gives the cosine value of all acute angles 1, 2, 3, . . can be seen here.
Let us now see a solved example:
Solved example 30.3
Given that cos 32 = 0.8480. Find the hypotenuse AC of the ⊿ABC shown in fig.30.9(a) below:
Solution:
• In this problem, we are given both the opposite side and adjacent side of 32o.
• But we are given only the cosine of 32o. Sine is not given. So we will use the adjacent side.
1. Given that cos 32 = 0.8480
2. So in any right triangle with one angle 32, the ratio
adjacent side of that 32⁄hypotenuse of that triangle = 0.8480
3. So, in our present triangle, we can write:
AB⁄AC = 0.8480 ⟹ 6.4013⁄AC = 0.8480 ⟹ AC = 6.4013⁄0.8480 = 7.5487 m
Check:
Applying Pythagoras theorem we have:
AC2 = AB2 + BC2 ⟹ AC2 = 6.40132 + 42 ⟹ AB2 = 40.9766 + 16 ⟹ AB2 = 56.9766
⟹ AB = √56.9766 = 7.5483 m
Solved example 30.4
Using the details given for ⊿PQR in fig.30.9(b), find the hypotenuse of ⊿UVW. Assume that tables are not available for finding cosine values.
Solution:
1. First we calculate the hypotenuse PR of PQR. For that, we can use Pythagoras theorem:
PR2 = PQ 2 + QR2 ⟹ PR2 = 1.34902 + 22 ⟹ PR2 = 1.8198 + 4 ⟹ PR2 = 5.8198
⟹ PR = √5.8198 = 2.4124 m
2. So in ⊿PQR, cos 56 = adjacent side of 56⁄hypotenuse
= PQ⁄PR = 1.3490⁄2.4124
3. In ⊿UVW, we can write:
cos 56 = adjacent side of 56⁄hypotenuse = UW⁄VW = 1.6863⁄VW
4. Coine of an acute angle will be the same in any right triangle. So from (2) and (3) we get:
1.3490⁄2.4124 = 1.6863⁄VW = ⟹ VW = (1.6863 × 2.4124)⁄1.3490 = 3.0155
Experiment 3:
1. A person goes to the ⊿ABC in fig.30.6(a) below. He takes up position at vertex A
[Note that, fig.30.6 below is the same fig.30.2 that we saw at the beginning of the previous section, for Experiment 1]
 |
| Fig.30.6 |
• 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: adjacent side of 37⁄hypotenuse of the triangle
[Note that, in experiment 1, we used 'opposite side of 37' in the numerator]
• We get: AB⁄AC = 5.4242⁄6.7918 = 0.7986
• Note down this value. We will continue our experiment:
5. The person now goes to ⊿MNO in fig.30.6(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: adjacent side of 37⁄hypotenuse of the triangle
• We get: MN⁄OM = 7.7311⁄9.6804 = 0.7986
■ 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.30.6(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: adjacent side of 37⁄hypotenuse of the triangle
• We get: PR⁄PQ = 6.9884⁄8.7504 = 0.7986
■ 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.30.6(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: adjacent side of 37⁄hypotenuse of the triangle
• We get: VW⁄UV = 5.0312⁄6.2998 = 0.7986
■ 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: adjacent side of 37⁄hypotenuse of the triangle
• In all of them we got the same value 0.7986
■ So we can conclude:
• Whenever we get a right triangle with one acute angle 37o, we can write:
adjacent side of that 37⁄hypotenuse of that triangle = 0.7986
• We don't even have to do any calculations with the sides
• That means, adjacent side of 37⁄hypotenuse of the triangle is a constant 0.7986
• Later in this chapter, we will see the reason for this 'constant nature'. In fact, corresponding to 37o, the value 0.7986 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 4:1. A person goes to the ⊿ABC in fig.30.7(a) below. He takes up position at vertex A
 |
| Fig.30.7 |
• 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: adjacent side of 68⁄hypotenuse of the triangle
• We get: AB⁄AC = 4.1578⁄11.0991 = 0.3746
• 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: adjacent side of 68⁄hypotenuse of the triangle
• We get: MN⁄OM = 2.0687⁄5.5222 = 0.3746
■ 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: adjacent side of 68⁄hypotenuse of the triangle
• We get: PR⁄PQ = 3.2926⁄8.7894 = 0.3746
■ 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: adjacent side of 68⁄hypotenuse of the triangle
• We get: UW⁄UV = 6.2606⁄16.7123 = 0.3746
■ 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: adjacent side of 68⁄hypotenuse of the triangle
• In all of them we got the same value 0.3746
■ So we can conclude:
• Whenever we get a right triangle with one acute angle 68o, we can write:
adjacent side of that 68⁄hypotenuse of that triangle = 0.3746
• We don't even have to do any calculations with the sides
• That means, adjacent side of 68⁄hypotenuse of the triangle is a constant 0.3746
• Later in this chapter, we will see the reason for this 'constant nature'. In fact, corresponding to 68o, the value 0.3746 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. 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: adjacent side⁄hypotenuse
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: adjacent side of θ⁄hypotenuse of the 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.30.8 below:
 |
| Fig.30.8 |
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 second and the last ratios [Note that, in the discussion on sine, we took the first and the last ratios]. We get:
• AB⁄PR = AC⁄PQ This can be written as:
• AB⁄AC = PR ⁄PQ
• That is: [adjacent side of θ⁄hypotenuse in ⊿ABC] = [adjacent side of θ⁄hypotenuse in ⊿PQR] =
• Thus we find that the ratio is same in the two triangles.
7. So it is clear that the ratio adjacent side of θ⁄hypotenuse will be a unique value. We can write:• AB⁄PR = AC⁄PQ This can be written as:
• AB⁄AC = PR ⁄PQ
• That is: [adjacent side of θ⁄hypotenuse in ⊿ABC] = [adjacent side of θ⁄hypotenuse in ⊿PQR] =
• Thus we find that the ratio is same in the two triangles.
• adjacent side of 1o⁄hypotenuse = a unique value for 1o. In the standard tables it is given as 0.9998
• adjacent side of 2o⁄hypotenuse = a unique value for 2o. In the standard tables it is given as 0.9994
• adjacent side of 3o⁄hypotenuse = a unique value for 3o. In the standard tables it is given as 0.9986
so on...
■ In general, for any acute angle θ, whenever we want the ratio adjacent side of θ⁄hypotenuse , 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: 'Cosine'.
• adjacent side of θ⁄hypotenuse is denoted as: 'cosine of θ'. It is abbreviated as 'cos θ'.
• So, from experiment 3, we have: cos 37 = 0.7986
• from experiment 4, we have: cos 68 = 0.3746
• A standard table that gives the cosine value of all acute angles 1, 2, 3, . . can be seen here.
Let us now see a solved example:
Solved example 30.3
Given that cos 32 = 0.8480. Find the hypotenuse AC of the ⊿ABC shown in fig.30.9(a) below:
 |
| Fig.30.9 |
• In this problem, we are given both the opposite side and adjacent side of 32o.
• But we are given only the cosine of 32o. Sine is not given. So we will use the adjacent side.
1. Given that cos 32 = 0.8480
2. So in any right triangle with one angle 32, the ratio
adjacent side of that 32⁄hypotenuse of that triangle = 0.8480
3. So, in our present triangle, we can write:
AB⁄AC = 0.8480 ⟹ 6.4013⁄AC = 0.8480 ⟹ AC = 6.4013⁄0.8480 = 7.5487 m
Check:
Applying Pythagoras theorem we have:
AC2 = AB2 + BC2 ⟹ AC2 = 6.40132 + 42 ⟹ AB2 = 40.9766 + 16 ⟹ AB2 = 56.9766
⟹ AB = √56.9766 = 7.5483 m
Solved example 30.4
Using the details given for ⊿PQR in fig.30.9(b), find the hypotenuse of ⊿UVW. Assume that tables are not available for finding cosine values.
Solution:
1. First we calculate the hypotenuse PR of PQR. For that, we can use Pythagoras theorem:
PR2 = PQ 2 + QR2 ⟹ PR2 = 1.34902 + 22 ⟹ PR2 = 1.8198 + 4 ⟹ PR2 = 5.8198
⟹ PR = √5.8198 = 2.4124 m
2. So in ⊿PQR, cos 56 = adjacent side of 56⁄hypotenuse
= PQ⁄PR = 1.3490⁄2.4124
3. In ⊿UVW, we can write:
cos 56 = adjacent side of 56⁄hypotenuse = UW⁄VW = 1.6863⁄VW
4. Coine of an acute angle will be the same in any right triangle. So from (2) and (3) we get:
1.3490⁄2.4124 = 1.6863⁄VW = ⟹ VW = (1.6863 × 2.4124)⁄1.3490 = 3.0155
So we have completed the discussion on cosine. In the next section we will see tangent.
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