In the previous section we completed a discussion on second degree equations. In this chapter we will learn the basics of Trigonometry.
Let us recall some of the important properties of Right angled triangles:
• In a right triangle, one angle is a right angle. That is., a 90o angle. See fig.30.1 below:
• The side opposite to this 90o angle is the hypotenuse
• The sum of the two non-right angles will be 90 [∵ sum of the three interior angles of any triangle is 180. So (1st non-right angle + 2nd non-right angle) + 90 = 180]
• So each of the two non-right angles will be less than 90o. That is., each of the two non-right angles will be acute angles
• If we know any one of these two acute angles, we can easily calculate the other. This is because, [known acute angle + unknown acute angle + 90] = 180
Based on these information, let us do an experiment.
Experiment 1:
1. A person goes to the ⊿ABC in fig.30.2(a) below. He takes up position at vertex A
[Some information about the triangles in the fig. can be seen here]
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⁄hypotenuse of the triangle
• We get: BC⁄AC = 4.0874⁄6.7918 = 0.6018
• Note down this value. We will continue our experiment:
5. The person now goes to ⊿MNO in fig.30.2(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⁄hypotenuse of the triangle
• We get: ON⁄OM = 5.8258⁄9.6804 = 0.6018
■ 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.30.2(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⁄hypotenuse of the triangle
• We get: RQ⁄PQ = 5.2661⁄8.7504 = 0.6018
■ 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.30.2(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⁄hypotenuse of the triangle
• We get: UW⁄UV = 3.7913⁄6.2998 = 0.6018
■ This is the same value that we obtained in (4), (8) & (12)
• Experiment 1 is now complete
• All of them had one acute angle of 37o
• In all of them we took the ratio: opposite side of 37⁄hypotenuse of the triangle
• In all of them we got the same value 0.6018
■ So we can conclude:
• Whenever we get a right triangle with one acute angle 37o, we can write:
opposite side of that 37⁄hypotenuse of that triangle = 0.6018
• We don't even have to do any calculations with the sides
• That means, opposite side of 37⁄hypotenuse of the triangle is a constant 0.6018
• Later in this section, we will see the reason for this 'constant nature'. In fact, corresponding to 37o, the value 0.6018 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.30.3(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⁄hypotenuse of the triangle
• We get: BC⁄AC = 10.2909⁄11.0991 = 0.9272
• Note down this value. We will continue our experiment:
5. The person now goes to ⊿MNO in fig.30.3(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⁄hypotenuse of the triangle
• We get: ON⁄OM = 5.1201⁄5.5222 = 0.9272
■ 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.30.3(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⁄hypotenuse of the triangle
• We get: RQ⁄PQ = 8.1494⁄8.7894 = 0.9272
■ 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.30.3(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⁄hypotenuse of the triangle
• We get: VW⁄UV = 15.4954⁄16.7123 = 0.9272
■ This is the same value that we obtained in (4), (8) & (12)
• Experiment 2 is now complete
• All of them had one acute angle of 68o
• In all of them we took the ratio: opposite side of 68⁄hypotenuse of the triangle
• In all of them we got the same value 0.9272
■ So we can conclude:
• Whenever we get a right triangle with one acute angle 68o, we can write:
opposite side of that 68⁄hypotenuse of that triangle = 0.9272
• We don't even have to do any calculations with the sides
• That means, opposite side of 68⁄hypotenuse of the triangle is a constant 0.9272
• Later in this section, we will see the reason for this 'constant nature'. In fact, corresponding to 68o, the value 0.9272 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: opposite 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: opposite 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.30.4 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⁄hypotenuse = a unique value for 1o. In the standard tables it is given as 0.0174
• opposite side of 2o⁄hypotenuse = a unique value for 2o. In the standard tables it is given as 0.0349
• opposite side of 3o⁄hypotenuse = a unique value for 3o. In the standard tables it is given as 0.0523
so on...
■ In general, for any acute angle θ, whenever we want the ratio opposite 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: 'Sine'.
• opposite side of θ⁄hypotenuse is denoted as: 'sine of θ'. It is abbreviated as 'sin θ'.
• So, from experiment 1, we have: sin 37 = 0.6018
• from experiment 2, we have: sin 68 = 0.9261
• A standard table that gives the sine value of all acute angles 1, 2, 3, . . can be seen here.
Let us now see a solved example:
Solved example 30.1
Given that sin 32 = 0.5299. Find the hypotenuse AC of the ⊿ABC shown in fig.30.5(a) below:
Solution:
1. Given that sin 32 = 0.5299.
2. So in any right triangle with one angle 32, the ratio
opposite side of that 32⁄hypotenuse of that triangle = 0.5299
3. So, in our present triangle, we can write:
BC⁄AC = 0.5299 ⟹ 4⁄AC = 0.5299 ⟹ AC = 4⁄0.5299= 7.5486 m
Solved example 30.2
Using the details given for ⊿PQR in fig.30.5(b), find the hypotenuse of ⊿UVW. Assume that tables are not available for finding sine values.
Solution:
1. First we calculate the hypotenuse PR of PQR. For that, we can use Pythagoras theorem:
PR2 = PQ2 + QR2 ⟹ PR2 = 1.34902 + 22 ⟹ PR2 = 1.8198 + 4 ⟹ PR2 = 5.8198
⟹ PR = √5.8198 = 2.4124 m
2. So in ⊿PQR, sin 56 = opposite side of 56⁄hypotenuse
= QR⁄PR = 2⁄2.4124
3. In ⊿UVW, we can write:
sin 56 = opposite side of 56⁄hypotenuse = UV⁄VW = 2.5⁄VW
4. Sine of an acute angle will be the same in any right triangle. So from (2) and (3) we get:
2⁄2.4124 = 2.5⁄VW = ⟹ VW = (2.5 × 2.4124)⁄2 = 3.0155
Let us recall some of the important properties of Right angled triangles:
• In a right triangle, one angle is a right angle. That is., a 90o angle. See fig.30.1 below:
 |
| Fig.30.1 |
• The sum of the two non-right angles will be 90 [∵ sum of the three interior angles of any triangle is 180. So (1st non-right angle + 2nd non-right angle) + 90 = 180]
• So each of the two non-right angles will be less than 90o. That is., each of the two non-right angles will be acute angles
• If we know any one of these two acute angles, we can easily calculate the other. This is because, [known acute angle + unknown acute angle + 90] = 180
Based on these information, let us do an experiment.
Experiment 1:
1. A person goes to the ⊿ABC in fig.30.2(a) below. He takes up position at vertex A
[Some information about the triangles in the fig. can be seen here]
 |
| Fig.30.2 |
• 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⁄hypotenuse of the triangle
• We get: BC⁄AC = 4.0874⁄6.7918 = 0.6018
• Note down this value. We will continue our experiment:
5. The person now goes to ⊿MNO in fig.30.2(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⁄hypotenuse of the triangle
• We get: ON⁄OM = 5.8258⁄9.6804 = 0.6018
■ 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.30.2(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⁄hypotenuse of the triangle
• We get: RQ⁄PQ = 5.2661⁄8.7504 = 0.6018
■ 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.30.2(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⁄hypotenuse of the triangle
• We get: UW⁄UV = 3.7913⁄6.2998 = 0.6018
■ This is the same value that we obtained in (4), (8) & (12)
• Experiment 1 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⁄hypotenuse of the triangle
• In all of them we got the same value 0.6018
■ So we can conclude:
• Whenever we get a right triangle with one acute angle 37o, we can write:
opposite side of that 37⁄hypotenuse of that triangle = 0.6018
• We don't even have to do any calculations with the sides
• That means, opposite side of 37⁄hypotenuse of the triangle is a constant 0.6018
• Later in this section, we will see the reason for this 'constant nature'. In fact, corresponding to 37o, the value 0.6018 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 2:1. A person goes to the ⊿ABC in fig.30.3(a) below. He takes up position at vertex A
 |
| Fig.30.3 |
• 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⁄hypotenuse of the triangle
• We get: BC⁄AC = 10.2909⁄11.0991 = 0.9272
• Note down this value. We will continue our experiment:
5. The person now goes to ⊿MNO in fig.30.3(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⁄hypotenuse of the triangle
• We get: ON⁄OM = 5.1201⁄5.5222 = 0.9272
■ 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.30.3(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⁄hypotenuse of the triangle
• We get: RQ⁄PQ = 8.1494⁄8.7894 = 0.9272
■ 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.30.3(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⁄hypotenuse of the triangle
• We get: VW⁄UV = 15.4954⁄16.7123 = 0.9272
■ This is the same value that we obtained in (4), (8) & (12)
• Experiment 2 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⁄hypotenuse of the triangle
• In all of them we got the same value 0.9272
■ So we can conclude:
• Whenever we get a right triangle with one acute angle 68o, we can write:
opposite side of that 68⁄hypotenuse of that triangle = 0.9272
• We don't even have to do any calculations with the sides
• That means, opposite side of 68⁄hypotenuse of the triangle is a constant 0.9272
• Later in this section, we will see the reason for this 'constant nature'. In fact, corresponding to 68o, the value 0.9272 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: opposite 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: opposite 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.30.4 below:
 |
| Fig.30.4 |
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 last ratios. We get:
• BC⁄QR = AC⁄PQ This can be written as:
• BC⁄AC = QR ⁄PQ
• That is: [opposite side of θ⁄hypotenuse in ⊿ABC] = [opposite 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 opposite side of θ⁄hypotenuse will be a unique value. We can write:• BC⁄QR = AC⁄PQ This can be written as:
• BC⁄AC = QR ⁄PQ
• That is: [opposite side of θ⁄hypotenuse in ⊿ABC] = [opposite side of θ⁄hypotenuse in ⊿PQR] =
• Thus we find that the ratio is same in the two triangles.
• opposite side of 1o⁄hypotenuse = a unique value for 1o. In the standard tables it is given as 0.0174
• opposite side of 2o⁄hypotenuse = a unique value for 2o. In the standard tables it is given as 0.0349
• opposite side of 3o⁄hypotenuse = a unique value for 3o. In the standard tables it is given as 0.0523
so on...
■ In general, for any acute angle θ, whenever we want the ratio opposite 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: 'Sine'.
• opposite side of θ⁄hypotenuse is denoted as: 'sine of θ'. It is abbreviated as 'sin θ'.
• So, from experiment 1, we have: sin 37 = 0.6018
• from experiment 2, we have: sin 68 = 0.9261
• A standard table that gives the sine value of all acute angles 1, 2, 3, . . can be seen here.
Let us now see a solved example:
Solved example 30.1
Given that sin 32 = 0.5299. Find the hypotenuse AC of the ⊿ABC shown in fig.30.5(a) below:
 |
| Fig.30.5 |
1. Given that sin 32 = 0.5299.
2. So in any right triangle with one angle 32, the ratio
opposite side of that 32⁄hypotenuse of that triangle = 0.5299
3. So, in our present triangle, we can write:
BC⁄AC = 0.5299 ⟹ 4⁄AC = 0.5299 ⟹ AC = 4⁄0.5299= 7.5486 m
Solved example 30.2
Using the details given for ⊿PQR in fig.30.5(b), find the hypotenuse of ⊿UVW. Assume that tables are not available for finding sine values.
Solution:
1. First we calculate the hypotenuse PR of PQR. For that, we can use Pythagoras theorem:
PR2 = PQ2 + QR2 ⟹ PR2 = 1.34902 + 22 ⟹ PR2 = 1.8198 + 4 ⟹ PR2 = 5.8198
⟹ PR = √5.8198 = 2.4124 m
2. So in ⊿PQR, sin 56 = opposite side of 56⁄hypotenuse
= QR⁄PR = 2⁄2.4124
3. In ⊿UVW, we can write:
sin 56 = opposite side of 56⁄hypotenuse = UV⁄VW = 2.5⁄VW
4. Sine of an acute angle will be the same in any right triangle. So from (2) and (3) we get:
2⁄2.4124 = 2.5⁄VW = ⟹ VW = (2.5 × 2.4124)⁄2 = 3.0155
So we have completed the discussion on sine. In the next section we will see cosine.
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