In the previous section we saw some solved examples on 30o, 60o and 45o right triangles. In this section, we will see some applications of trigonometry related to circles.
• Consider the arc ABC in fig.30.23(a) below. It has a central angle of θo. Radius of the circle is 'r' cm.
• Line segment AC which joins the ends of the arc ABC is a chord of the circle.
• We have seen how to calculate the length of the arc ABC in previous classes. See details here.
• We have also seen how to calculate the length of chord AC. See details here. In this section we will see another method using Trigonometry.
1. Consider the ΔAOC in fig.30.23(b). Drop a perpendicular OD from the vertex O onto the side AC. This perpendicular will bisect AC. So AD = CD. The reason can be seen here.
2. Also, the perpendicular OD will bisect ∠AOC. So ∠AOD = ∠COD = θ⁄2 .
3. Now we have two right triangles: ⊿AOD and ⊿COD. Consider any one of them. Let us consider ⊿AOD. We get: sin θ⁄2 = AD⁄AO = AD⁄r ⟹ AD = r × sin θ⁄2. = r sin θ⁄2.
4. But AD = CD. That is., AD is half of the chord AC
• So the full length of the chord AC = 2 × r sin θ⁄2. = 2r sin θ⁄2.
• In this way we can find the length of any chord if the central angle and the radius of circle are known. We can write it in the form of a theorem. We will write it in steps:
Theorem 30.1:
1. Consider any chord in a circle of radius r
2. Let the chord make a central angle of θ⁄2
3. Find the sine of θ⁄2
4. Multiply this sine by 2r
5. The product will be the length of the chord.
That is., Length of the chord = 2r sin θ⁄2
An example:
Find the length of a chord whose central angle is 70. Radius of the circle is 3 cm
Solution:
1. Given: r = 3 cm, θ = 70o
2. Then length of the chord = 2r sin θ⁄2 = 2 × 3 × sin 70⁄2 = 6 sin 35 = 6 × 0.5736 = 3.4414 cm
• A diagram of this problem drawn with a computer program gives the same result. It is shown below:
• Consider any three points on a plane sheet of paper. If those three points are not on a line, we can draw a circle passing through all those three points.
• Also, a triangle can be formed using those three points. That means, we can draw a circle passing through the three vertices of any given triangle. It is called the circumcircle of the triangle. We learned all those details here.
• Consider the triangle and it's circle shown in the fig.30.24(a) below:
• The angle at vertex A is given. The radius of the circle is also given. We want to find the length of the chord BC. Is it possible? Let us try:
1. Consider the minor arc BC in fig.30.24(b). Our required chord joins the two end points of this minor arc BC
2. The minor arc BC makes an angle Ao on the alternate arc BAC. So the central angle of minor arc BC will be (2A)o. Details here.
3. So the central angle of the chord BC is also (2A)o.
4. Once we get the central angle of chord BC, we can easily find it's length. For that, we use theorem 30.1 that we saw above.
• Length of chord BC = 2r sin θ⁄2 = 2r sin (2A)⁄2 = 2r sin A
5. Note that, BC is the opposite side of vertex A. If we use another vertex B, we will get the length of chord AC. That is.,
• Length of chord AC = 2r sin B
• Similarly, Length of chord AB = 2r sin C
Now we have to check whether this formula is applicable to all triangles.
If an angle of a triangle is obtuse, we must subtract it from 180, to use in the formula.
An example:
Using the sine and cosine tables, and if needed, a calculator, do the following problems:
Solution:
1. Let us name the triangle as ΔABC. This is shown in fig(b). Let the radius of the circumcircle be 'r' cm
2. Then we get: AB = 2r sin C ⟹ 4 = 2r × sin 70 ⟹ 4 = 2r × 0.9397 ⟹ 2r = 4⁄0.9397 = 4.26 cm
3. But 2r is the diameter. So we can write:
• Diameter of the circumcircle = 4.26 cm
Solved example 3.17
A circle is to be drawn, passing through the ends of a line 5 cm long. This line should subtend an angle of 80o on one side. What should be the radius of the circle?
Solution:
Consider the rough sketch in fig.30.28 below
1. Let us name the triangle as ΔABC. Let the radius of the circumcircle be 'r' cm
2. Then we get: AB = 2r sin C ⟹ 5 = 2r × sin 80 ⟹ 5 = 2r × 0.9848
⟹ 2.5 = r × 0.9848 ⟹ r = 2.5⁄0.9848 = 2.538 cm
3. So we can write:
• Radius of the circumcircle = 2.538 cm
Solved example 30.18
A part of a circle is shown in fig.30.29(a) below. What is the radius of the circle?
Solution:
1. The possible full circle is shown in fig(b). We have to find the radius 'r' of this circle
2. Consider the minor arc ADC. It subtends ∠ABC on the alternate arc. Now we have a cyclic quadrilateral ABCD
3. In the cyclic quadrilateral, ∠ADC + ∠ABC = 180o. So ∠ABC = 180 - 140 = 40o
4. Consider ΔABC. We get: AC = 2r sin B ⟹ 8 = 2r sin 40 ⟹ 8 = 2r × 0.6428
⟹ 2r = 8⁄0.6428 ⟹ r = 4⁄0.6428 = 6.222 cm
Another method:
1. In fig(b), ∠ADC is an obtuse angle. So, to relate it with the length of chord AC, we must subtract it from 180o. (see fig.30.25 above)
2. We get: AC = 2r sin (180-140) ⟹ 8 = 2r sin 40 ⟹ 8 = 2r × 0.6428
⟹ 2r = 8⁄0.6428 ⟹ r = 4⁄0.6428 = 6.222 cm
Solved example 30.19
Draw the circle and the triangle shown in fig.30.30(a) in your note book and explain how it was drawn. Calculate the lengths of all three sides
Solution:
1. In fig(a) we have a circle of diameter 5 cm, and a triangle whose two angles are given
2. The third angle will be obviously [180-(45+65)] = [180 - 110] = 70o.
3. The given triangle is named as ΔABC in fig(b). Two more details are also added in this fig(b):
• The radius of the circle is 2.5 cm
• The minor arc AB subtends an angle of 70o on the alternate arc. So the central angle ∠AOB of this minor arc will be equal to 70 × 2 = 140o
• With these details we can do the construction. The steps are shown in the fig.30.31 below:
Step 1: • With any convenient point 'O' as centre, draw a circle of radius 2.5 cm
• At 'O', draw two lines with an angle of 140o between them
• Name the points of intersection of the lines with the circle as 'A' and 'B'
• These are shown in fig.30.31(a) above
Step 2: • Draw line AB. This is shown in fig(b)
Step 3: • At A, draw a second line at an angle of 45o with AB
• At B, draw a third line at an angle of 65o with AB
• These two lines will intersect at a point. This point will lie on the circle. Name this point as 'C'.
• Thus the construction is complete
Part 2: In this part, we have to calculate the sides of the ΔABC
1. AB = 2r sin C = 2×2.5×sin 70 = 2×2.5×0.9397 = 4.6985 cm
2. AC = 2r sin B = 2×2.5×sin 65 = 2×2.5×0.9063 = 4.5315 cm
3. BC = 2r sin A = 2×2.5×sin 45 = 2×2.5×0.7071 = 3.5355 cm
Solved example 30.20
A triangle is made by drawing angles of 50o and 65o at the ends of a 5 cm long line. Calculate it's area
Solution:
The given data is shown in fig.30.32(a) below:
1. Let us add some more details. The triangle is named as ΔABC. The third angle at C will be [180 - (50+65)] = [180-115] = 65o. This is shown in fig(b)
2. So two angles are 65o. It is an isosceles triangle. Sides opposite the equal angles are equal.
We get: AB = AC = 5 cm
3. Drop a perpendicular CD from the vertex C to the side AB. Consider the right triangle ⊿ADC.
We have: sin 50 = CD⁄AC ⟹ 0.7660 = CD⁄5 ⟹ CD = 0.7660 × 5 = 3.83 cm
4. Area of ΔABC = 1⁄2 × base × altitude = 1⁄2 × 5 × 3.83 = 9.58 cm2
• Consider the arc ABC in fig.30.23(a) below. It has a central angle of θo. Radius of the circle is 'r' cm.
• Line segment AC which joins the ends of the arc ABC is a chord of the circle.
• We have seen how to calculate the length of the arc ABC in previous classes. See details here.
• We have also seen how to calculate the length of chord AC. See details here. In this section we will see another method using Trigonometry.
1. Consider the ΔAOC in fig.30.23(b). Drop a perpendicular OD from the vertex O onto the side AC. This perpendicular will bisect AC. So AD = CD. The reason can be seen here.
2. Also, the perpendicular OD will bisect ∠AOC. So ∠AOD = ∠COD = θ⁄2 .
3. Now we have two right triangles: ⊿AOD and ⊿COD. Consider any one of them. Let us consider ⊿AOD. We get: sin θ⁄2 = AD⁄AO = AD⁄r ⟹ AD = r × sin θ⁄2. = r sin θ⁄2.
4. But AD = CD. That is., AD is half of the chord AC
• So the full length of the chord AC = 2 × r sin θ⁄2. = 2r sin θ⁄2.
• In this way we can find the length of any chord if the central angle and the radius of circle are known. We can write it in the form of a theorem. We will write it in steps:
Theorem 30.1:
1. Consider any chord in a circle of radius r
2. Let the chord make a central angle of θ⁄2
3. Find the sine of θ⁄2
4. Multiply this sine by 2r
5. The product will be the length of the chord.
That is., Length of the chord = 2r sin θ⁄2
An example:
Find the length of a chord whose central angle is 70. Radius of the circle is 3 cm
Solution:
1. Given: r = 3 cm, θ = 70o
2. Then length of the chord = 2r sin θ⁄2 = 2 × 3 × sin 70⁄2 = 6 sin 35 = 6 × 0.5736 = 3.4414 cm
• A diagram of this problem drawn with a computer program gives the same result. It is shown below:
Another application:
• Also, a triangle can be formed using those three points. That means, we can draw a circle passing through the three vertices of any given triangle. It is called the circumcircle of the triangle. We learned all those details here.
• Consider the triangle and it's circle shown in the fig.30.24(a) below:
1. Consider the minor arc BC in fig.30.24(b). Our required chord joins the two end points of this minor arc BC
2. The minor arc BC makes an angle Ao on the alternate arc BAC. So the central angle of minor arc BC will be (2A)o. Details here.
3. So the central angle of the chord BC is also (2A)o.
4. Once we get the central angle of chord BC, we can easily find it's length. For that, we use theorem 30.1 that we saw above.
• Length of chord BC = 2r sin θ⁄2 = 2r sin (2A)⁄2 = 2r sin A
5. Note that, BC is the opposite side of vertex A. If we use another vertex B, we will get the length of chord AC. That is.,
• Length of chord AC = 2r sin B
• Similarly, Length of chord AB = 2r sin C
1. Consider the circumcircle of ΔABC in fig.30.25(a) below.
•The centre of the circle is outside ΔABC. Note that, in the previous fig.30.24(a), the centre is inside ΔABC
2. The arc BC is now a major arc. This is shown in red colour in fig(b). The central angle of this arc is (2A)o
3. Now consider fig(c). We have, ∠COB = (360-2A)
4. Drop a perpendicular OD from O onto the side BC. This will bisect BC and also ∠COB
5. So we get: ∠COD = ∠BOD = COB⁄2 = (360-2A)⁄2 = (180-A)⁄2
6. In ⊿OCD, sin (180-A)⁄2 = CD⁄r ⟹ CD = r sin (180-A)⁄2
7. So BC = 2 × CD = 2r sin (180-A)⁄2.
• So we can write:If an angle of a triangle is obtuse, we must subtract it from 180, to use in the formula.
In fig.30.26 below, length of side AB is given. Find the lengths of the other two sides
Solution:
We have: AB = 2r sin C ⟹ 6 = 2r sin 80 ⟹ 6 = 2r × 0.9848 ⟹ 2r = 6⁄0.9848
⟹ 2r = 6.0926 cm
2. Now, BC = 2r sin A
Substituting the values of '2r' and 'A', we get: BC = 6.0926 × sin 60
⟹ BC = 6.0926 × 0.8660 = 5.28 cm
3. Similarly, AC = 2r sin B
Substituting the values of '2r' and 'B', we get: AC = 6.0926 × sin 40
⟹ BC = 6.0926 × 0.6428 = 3.92 cm
Solved example 3.16
A triangle and it's circumcircle are shown in the fig.30.27(a) below. Calculate the diameter of the circle
Solution:
1. Let us name the triangle as ΔABC. This is shown in fig(b). Let the radius of the circumcircle be 'r' cm
2. Then we get: AB = 2r sin C ⟹ 4 = 2r × sin 70 ⟹ 4 = 2r × 0.9397 ⟹ 2r = 4⁄0.9397 = 4.26 cm
3. But 2r is the diameter. So we can write:
• Diameter of the circumcircle = 4.26 cm
Solved example 3.17
A circle is to be drawn, passing through the ends of a line 5 cm long. This line should subtend an angle of 80o on one side. What should be the radius of the circle?
Solution:
Consider the rough sketch in fig.30.28 below
2. Then we get: AB = 2r sin C ⟹ 5 = 2r × sin 80 ⟹ 5 = 2r × 0.9848
⟹ 2.5 = r × 0.9848 ⟹ r = 2.5⁄0.9848 = 2.538 cm
3. So we can write:
• Radius of the circumcircle = 2.538 cm
Solved example 30.18
A part of a circle is shown in fig.30.29(a) below. What is the radius of the circle?
1. The possible full circle is shown in fig(b). We have to find the radius 'r' of this circle
2. Consider the minor arc ADC. It subtends ∠ABC on the alternate arc. Now we have a cyclic quadrilateral ABCD
3. In the cyclic quadrilateral, ∠ADC + ∠ABC = 180o. So ∠ABC = 180 - 140 = 40o
4. Consider ΔABC. We get: AC = 2r sin B ⟹ 8 = 2r sin 40 ⟹ 8 = 2r × 0.6428
⟹ 2r = 8⁄0.6428 ⟹ r = 4⁄0.6428 = 6.222 cm
Another method:
1. In fig(b), ∠ADC is an obtuse angle. So, to relate it with the length of chord AC, we must subtract it from 180o. (see fig.30.25 above)
2. We get: AC = 2r sin (180-140) ⟹ 8 = 2r sin 40 ⟹ 8 = 2r × 0.6428
⟹ 2r = 8⁄0.6428 ⟹ r = 4⁄0.6428 = 6.222 cm
Solved example 30.19
Draw the circle and the triangle shown in fig.30.30(a) in your note book and explain how it was drawn. Calculate the lengths of all three sides
1. In fig(a) we have a circle of diameter 5 cm, and a triangle whose two angles are given
2. The third angle will be obviously [180-(45+65)] = [180 - 110] = 70o.
3. The given triangle is named as ΔABC in fig(b). Two more details are also added in this fig(b):
• The radius of the circle is 2.5 cm
• The minor arc AB subtends an angle of 70o on the alternate arc. So the central angle ∠AOB of this minor arc will be equal to 70 × 2 = 140o
• With these details we can do the construction. The steps are shown in the fig.30.31 below:
• At 'O', draw two lines with an angle of 140o between them
• Name the points of intersection of the lines with the circle as 'A' and 'B'
• These are shown in fig.30.31(a) above
Step 2: • Draw line AB. This is shown in fig(b)
Step 3: • At A, draw a second line at an angle of 45o with AB
• At B, draw a third line at an angle of 65o with AB
• These two lines will intersect at a point. This point will lie on the circle. Name this point as 'C'.
• Thus the construction is complete
Part 2: In this part, we have to calculate the sides of the ΔABC
1. AB = 2r sin C = 2×2.5×sin 70 = 2×2.5×0.9397 = 4.6985 cm
2. AC = 2r sin B = 2×2.5×sin 65 = 2×2.5×0.9063 = 4.5315 cm
3. BC = 2r sin A = 2×2.5×sin 45 = 2×2.5×0.7071 = 3.5355 cm
Solved example 30.20
A triangle is made by drawing angles of 50o and 65o at the ends of a 5 cm long line. Calculate it's area
Solution:
The given data is shown in fig.30.32(a) below:
2. So two angles are 65o. It is an isosceles triangle. Sides opposite the equal angles are equal.
We get: AB = AC = 5 cm
3. Drop a perpendicular CD from the vertex C to the side AB. Consider the right triangle ⊿ADC.
We have: sin 50 = CD⁄AC ⟹ 0.7660 = CD⁄5 ⟹ CD = 0.7660 × 5 = 3.83 cm
4. Area of ΔABC = 1⁄2 × base × altitude = 1⁄2 × 5 × 3.83 = 9.58 cm2
In the next section we will see problems related to the other trigonometric ratio tan.
No comments:
Post a Comment