In the previous section we saw some solved examples demonstrating the basic principles of trigonometry. In this section, we will see some interesting details about right triangles with 30o, 60o and 45o.
• First we will consider right triangle with 30o. When one of the acute angles in a right triangle is 30o, the other acute angle will be obviously 60o.
• This is very convenient for us. We can discuss 30o and 60o together.
Fig. 30.13(a) below shows a ⊿ABC. It is right angled at B. The angle at vertex C is 30o. So the angle at vertex A will be 60o.
1. Let the length of the hypotenuse AC be 's' cm.
2. Let us take the mirror image of this triangle. The mirror line is the altitude BC. This is shown in fig.(b).
3. Since BC is the mirror line, vertices B and C does not change. But vertex A has a corresponding point on the other side of the mirror line. Let us call it 'D'.
4. So now we have a new ΔADC. Note that it is not right angled. Let us write the details about this new ΔADC:
• DC is the mirror image of AC. So length of DC = s
• Angle at vertex C in ΔADC = (30 + 30) = 60o
• Since D is the mirror image of A, Angle at vertex D = 60o
• Thus in ADC, all the three angles are 60o. So it is an equilateral triangle
• In an equilateral triangle, all sides will be equal. So we get length of AD = s cm
5. Now, in the equilateral triangle ADC, CB is perpendicular to AD. So CB bisects AD. Thus we get: AB = BD = s⁄2
6. Inside the large ΔADC, consider ⊿ABC. It is a right triangle. So we can find BC by applying the Pythagoras theorem:
BC2 = AC2 - AB2 ⟹ BC2 = s2 - (s⁄2)2 ⟹ BC2 = [(4s2 - s2) ⁄4] = [(3s2) ⁄4]
⟹ BC = √[(3s2) ⁄4] = (√3)s⁄2
7. Now we have all the three sides of the original ⊿ABC:
• AB = s⁄2
• BC = (√3)s⁄2
• AC = s
8. We are in a position to apply the trigonometrical ratios:
• sin 30 = opposite side⁄hypotenuse = AB⁄AC = (s⁄2) ÷ s = (s⁄2) × (1⁄s) = 1⁄2 = 0.5
• sin 60 = opposite side⁄hypotenuse = BC⁄AC = [(√3)s⁄2] ÷ s = [(√3)s⁄2] × (1⁄s) = [√3⁄2] = 0.8660
• cos 30 = adjacent side⁄hypotenuse = BC⁄AC = [(√3)s⁄2] ÷ s = [(√3)s⁄2] × (1⁄s) = [√3⁄2] = 0.8660
• cos 60 = adjacent side⁄hypotenuse = AB⁄AC = (s⁄2) ÷ s = (s⁄2) × (1⁄s) = 1⁄2 = 0.5
• tan 30 = opposite side⁄adjacent side = AB⁄BC = (s⁄2) ÷ [(√3)s⁄2] = (s⁄2) × [2⁄(√3)s] = [1⁄√3] = 0.57735
• tan 60 = opposite side⁄adjacent side = BC⁄AB = [(√3)s⁄2] ÷ (s⁄2) = [(√3)s⁄2] × (2⁄s) = √3 = 1.73205
• Area of ΔADC = 1⁄2 × base × altitude = 1⁄2 × AD × BC
♦ AD = s
♦ Using ⊿ABC, we have already seen that BC = [(√3)s⁄2]
• So the area = 1⁄2 × s × [(√3)s⁄2] = [(√3)s2⁄4]
• This is the same formula shown in the video presentation
• Now we will consider right triangle with 45o. When one of the acute angles in a right triangle is 45o, the other acute angle will be obviously 45o.
• Fig.30.14 shows a ⊿ABC with acute angles 45o.
1. The base angles are equal. So it is an isosceles triangle. The sides opposite the equal angles are equal. So we have: AB = BC = s
2. It is also a right triangle. Applying Pythagoras theorem, we get:
AC2 = AB2 + BC2 ⟹ AC2 = s2 + s2 ⟹ AC2 = 2s2
So AC = √[2s2] = (√2)s
3. Now we have all the three sides of the original ⊿ABC:
• AB = s
• BC = s
• AC = (√2)s
4. We are in a position to apply the trigonometrical ratios:
• sin 45 = opposite side⁄hypotenuse = BC⁄AC = s ÷ (√2)s = s × (1⁄√2s) = 1⁄√2 = 0.7071
• cos 45 = adjacent side⁄hypotenuse = AB⁄AC = s ÷ (√2)s = s × (1⁄√2s) = 1⁄√2 = 0.7071
• tan 45 = opposite side⁄adjacent side = BC⁄AB = s ÷ s = 1
In the case of 45o right triangle, we can easily find the area. In fact we do not need to apply trigonometry for finding the area. Fig.30.14(b) above, shows the same ⊿ABC in another position. We can see that, base AB is 's' and altitude BC is also 's'. So we get:
• Area of ΔABC = 1⁄2 × base × altitude = 1⁄2 × s × s = s2⁄2
So we can write:
■ Area of an isosceles right triangle is always s2⁄2. Where s is the length of the equal sides.
Note that, for using this formula, the triangle must be isosceles and at the same time, right.
• First we will consider right triangle with 30o. When one of the acute angles in a right triangle is 30o, the other acute angle will be obviously 60o.
• This is very convenient for us. We can discuss 30o and 60o together.
Fig. 30.13(a) below shows a ⊿ABC. It is right angled at B. The angle at vertex C is 30o. So the angle at vertex A will be 60o.
 |
| Fig.30.13 |
2. Let us take the mirror image of this triangle. The mirror line is the altitude BC. This is shown in fig.(b).
3. Since BC is the mirror line, vertices B and C does not change. But vertex A has a corresponding point on the other side of the mirror line. Let us call it 'D'.
4. So now we have a new ΔADC. Note that it is not right angled. Let us write the details about this new ΔADC:
• DC is the mirror image of AC. So length of DC = s
• Angle at vertex C in ΔADC = (30 + 30) = 60o
• Since D is the mirror image of A, Angle at vertex D = 60o
• Thus in ADC, all the three angles are 60o. So it is an equilateral triangle
• In an equilateral triangle, all sides will be equal. So we get length of AD = s cm
5. Now, in the equilateral triangle ADC, CB is perpendicular to AD. So CB bisects AD. Thus we get: AB = BD = s⁄2
6. Inside the large ΔADC, consider ⊿ABC. It is a right triangle. So we can find BC by applying the Pythagoras theorem:
BC2 = AC2 - AB2 ⟹ BC2 = s2 - (s⁄2)2 ⟹ BC2 = [(4s2 - s2) ⁄4] = [(3s2) ⁄4]
⟹ BC = √[(3s2) ⁄4] = (√3)s⁄2
7. Now we have all the three sides of the original ⊿ABC:
• AB = s⁄2
• BC = (√3)s⁄2
• AC = s
8. We are in a position to apply the trigonometrical ratios:
• sin 30 = opposite side⁄hypotenuse = AB⁄AC = (s⁄2) ÷ s = (s⁄2) × (1⁄s) = 1⁄2 = 0.5
• sin 60 = opposite side⁄hypotenuse = BC⁄AC = [(√3)s⁄2] ÷ s = [(√3)s⁄2] × (1⁄s) = [√3⁄2] = 0.8660
• cos 30 = adjacent side⁄hypotenuse = BC⁄AC = [(√3)s⁄2] ÷ s = [(√3)s⁄2] × (1⁄s) = [√3⁄2] = 0.8660
• cos 60 = adjacent side⁄hypotenuse = AB⁄AC = (s⁄2) ÷ s = (s⁄2) × (1⁄s) = 1⁄2 = 0.5
• tan 30 = opposite side⁄adjacent side = AB⁄BC = (s⁄2) ÷ [(√3)s⁄2] = (s⁄2) × [2⁄(√3)s] = [1⁄√3] = 0.57735
• tan 60 = opposite side⁄adjacent side = BC⁄AB = [(√3)s⁄2] ÷ (s⁄2) = [(√3)s⁄2] × (2⁄s) = √3 = 1.73205
So now we know the trigonometric ratios of a 30o, 60o triangle. This is a good time to discuss about the 'area of equilateral triangles'. We have derived the formula in our earlier classes. It can be seen in the form of a video presentation here. Let us now derive it using trigonometry:
In fig.30.13(b) above, we have:• Area of ΔADC = 1⁄2 × base × altitude = 1⁄2 × AD × BC
♦ AD = s
♦ Using ⊿ABC, we have already seen that BC = [(√3)s⁄2]
• So the area = 1⁄2 × s × [(√3)s⁄2] = [(√3)s2⁄4]
• This is the same formula shown in the video presentation
• Now we will consider right triangle with 45o. When one of the acute angles in a right triangle is 45o, the other acute angle will be obviously 45o.
• Fig.30.14 shows a ⊿ABC with acute angles 45o.
 |
| Fig.30.14 |
2. It is also a right triangle. Applying Pythagoras theorem, we get:
AC2 = AB2 + BC2 ⟹ AC2 = s2 + s2 ⟹ AC2 = 2s2
So AC = √[2s2] = (√2)s
3. Now we have all the three sides of the original ⊿ABC:
• AB = s
• BC = s
• AC = (√2)s
4. We are in a position to apply the trigonometrical ratios:
• sin 45 = opposite side⁄hypotenuse = BC⁄AC = s ÷ (√2)s = s × (1⁄√2s) = 1⁄√2 = 0.7071
• cos 45 = adjacent side⁄hypotenuse = AB⁄AC = s ÷ (√2)s = s × (1⁄√2s) = 1⁄√2 = 0.7071
• tan 45 = opposite side⁄adjacent side = BC⁄AB = s ÷ s = 1
In the case of 45o right triangle, we can easily find the area. In fact we do not need to apply trigonometry for finding the area. Fig.30.14(b) above, shows the same ⊿ABC in another position. We can see that, base AB is 's' and altitude BC is also 's'. So we get:
• Area of ΔABC = 1⁄2 × base × altitude = 1⁄2 × s × s = s2⁄2
So we can write:
■ Area of an isosceles right triangle is always s2⁄2. Where s is the length of the equal sides.
Note that, for using this formula, the triangle must be isosceles and at the same time, right.
We have seen the trigonometric ratios of 30, 45 and 60 angles. Right triangles with these three angles are often encountered in scientific and engineering problems. So it is useful to remember them. They are given in a tabular form below:
| 30o | 60o | 45o | |
|---|---|---|---|
| Sine | 1⁄2 | √3⁄2 | 1⁄√2 |
| Cosine | √3⁄2 | 1⁄2 | 1⁄√2 |
| Tangent | 1⁄√3 | √3 | 1 |
Now we will see an interesting application of the above values:
1. Consider any 30o, 60o triangle. We have:
sin30 = opposite side of 30 in that triangle⁄hypotenuse of that triangle = 1⁄2
⟹ 2 × opposite side of 30 in that triangle = hypotenuse of that triangle
2. But 30o is the smallest angle in a 30o, 60o triangle. So the side opposite 30o will be the smallest side in a 30o, 60o triangle. Thus we can write:
■ In a 30o, 60o triangle, the hypotenuse is always twice the smallest side
Another result:
1. Consider any 30o, 60o triangle. We have:
tan 30 = opposite side of 30 in that triangle⁄adjacent side of 30 in that triangle = 1⁄√3
⟹ √3 × opposite side of 30 in that triangle = adjacent side of 30 in that triangle
2. But 30o is the smallest angle in a 30o, 60o triangle. So the side opposite 30o will be the smallest side in a 30o, 60o triangle.
• Also, the side adjacent to 30 will be the medium side in any 30o, 60o triangle. Thus we can write:
■ In a 30o, 60o triangle, the medium side is always √3 times the smallest side
Combining the above two results, we can write:
• If 'x' is the length of the smallest side in a 30o, 60o triangle, Then:
♦ hypotenuse = 2x
♦ medium side = (√3)x
• In other words, the ratio smallest side : medium side : hypotenuse in a 30o, 60o triangle is:
x : (√3)x : 2x
⟹ The ratio [smallest side : medium side : hypotenuse] in a 30o, 60o triangle is:
[1 : √3 : 2]
[Note that 2 > √3. So hypotenuse is indeed the longest side in a 30o, 60o triangle. In fact, the hypotenuse is the longest side in any right triangle. Because, it is opposite the largest angle (90o) in a right triangle]
sin 45 = opposite side of 45 in that triangle⁄hypotenuse of that triangle = 1⁄√2
⟹ √2 × opposite side of 45 in that triangle = hypotenuse of that triangle
2. But 45o is the smallest angle in a 45o isosceles triangle. So the side opposite 45o will be the smallest side in a 45o isosceles triangle. Thus we can write:
■ In a 45o isosceles triangle, the hypotenuse is always √2 times the smallest side
• But in a 45o isosceles triangle, there are two 'smallest sides'. The above result is applicable to both the 'smallest sides'.
So we can write:
• If 'x' is the length of the smallest side in a 45o isosceles triangle, Then:
♦ other smallest side = x
♦ hypotenuse = (√2)x
• In other words, the ratio smallest side : medium side : hypotenuse in a 45o isosceles triangle is:
x : x : (√2)x
⟹ The ratio [smallest side : medium side : hypotenuse] in a 45o isosceles triangle is:
[1 : 1 : √2]
[Note that √2 > 1. So hypotenuse is indeed the longest side in a 45o isosceles triangle]
• 45o isosceles triangle
■ On seeing the name 30o, 60o triangle we can immediately understand that it is a right triangle
■ On seeing the name 45o isosceles triangle we can immediately understand that it is a right triangle
The reader may write the reasons in his/her own notebooks
1. Consider any 30o, 60o triangle. We have:
sin30 = opposite side of 30 in that triangle⁄hypotenuse of that triangle = 1⁄2
⟹ 2 × opposite side of 30 in that triangle = hypotenuse of that triangle
2. But 30o is the smallest angle in a 30o, 60o triangle. So the side opposite 30o will be the smallest side in a 30o, 60o triangle. Thus we can write:
■ In a 30o, 60o triangle, the hypotenuse is always twice the smallest side
Another result:
1. Consider any 30o, 60o triangle. We have:
tan 30 = opposite side of 30 in that triangle⁄adjacent side of 30 in that triangle = 1⁄√3
⟹ √3 × opposite side of 30 in that triangle = adjacent side of 30 in that triangle
2. But 30o is the smallest angle in a 30o, 60o triangle. So the side opposite 30o will be the smallest side in a 30o, 60o triangle.
• Also, the side adjacent to 30 will be the medium side in any 30o, 60o triangle. Thus we can write:
■ In a 30o, 60o triangle, the medium side is always √3 times the smallest side
Combining the above two results, we can write:
• If 'x' is the length of the smallest side in a 30o, 60o triangle, Then:
♦ hypotenuse = 2x
♦ medium side = (√3)x
• In other words, the ratio smallest side : medium side : hypotenuse in a 30o, 60o triangle is:
x : (√3)x : 2x
⟹ The ratio [smallest side : medium side : hypotenuse] in a 30o, 60o triangle is:
[1 : √3 : 2]
[Note that 2 > √3. So hypotenuse is indeed the longest side in a 30o, 60o triangle. In fact, the hypotenuse is the longest side in any right triangle. Because, it is opposite the largest angle (90o) in a right triangle]
Next we will derive a similar result for the 45o isosceles triangle
1. Consider any 45o isosceles triangle. We have:sin 45 = opposite side of 45 in that triangle⁄hypotenuse of that triangle = 1⁄√2
⟹ √2 × opposite side of 45 in that triangle = hypotenuse of that triangle
2. But 45o is the smallest angle in a 45o isosceles triangle. So the side opposite 45o will be the smallest side in a 45o isosceles triangle. Thus we can write:
■ In a 45o isosceles triangle, the hypotenuse is always √2 times the smallest side
• But in a 45o isosceles triangle, there are two 'smallest sides'. The above result is applicable to both the 'smallest sides'.
So we can write:
• If 'x' is the length of the smallest side in a 45o isosceles triangle, Then:
♦ other smallest side = x
♦ hypotenuse = (√2)x
• In other words, the ratio smallest side : medium side : hypotenuse in a 45o isosceles triangle is:
x : x : (√2)x
⟹ The ratio [smallest side : medium side : hypotenuse] in a 45o isosceles triangle is:
[1 : 1 : √2]
[Note that √2 > 1. So hypotenuse is indeed the longest side in a 45o isosceles triangle]
In the above discussion, we used two names frequently:
• 30o, 60o triangle• 45o isosceles triangle
■ On seeing the name 30o, 60o triangle we can immediately understand that it is a right triangle
■ On seeing the name 45o isosceles triangle we can immediately understand that it is a right triangle
The reader may write the reasons in his/her own notebooks
Now we will see an example:
In the fig(a) below, we have a triangle whose angles are 45o, 30o and 105o.
It is not a right triangle. We want the ratio of it's sides. The procedure is as follows:
1. Let us name the given triangle as PQR
2. Drop a perpendicular RS from the vertex R on to the side PQ. This is shown in fig(b)
3. Now we have two right triangles: ⊿PSR and ⊿QSR
4. Consider ⊿PSR. We get ∠PRS = 45o.
Consider ⊿QSR. We get ∠QRS = 60o
5. So PSR is a 45o isosceles triangle
QSR is a 30o, 60o triangle
6. Consider ⊿PSR again. Let RS be 'x' cm.
• In a 45o isosceles triangle, we have: Hypotenuse = (√2) times the smallest side. In ⊿PSR, RS is one of the smallest sides because, it is opposite the smallest angle 45o
♦ So we can write: PR = (√2)×RS = (√2)x
Also in a 45o isosceles triangle, the smallest sides are equal.
♦ So we can write: PS = RS = x
These are shown in fig(c)
7. Consider ⊿QSR again
In a 30o, 60o triangle, we have: Hypotenuse = 2 times the smallest side. In ⊿QSR, RS is the smallest side because, it is opposite the smallest angle 30o
♦ So we can write: QR = 2×RS = 2x
Also in a 30o, 60o triangle, the medium side is (√3) times the smallest side. In ⊿QSR. SQ is the medium side because, it is opposite the medium angle 60o
♦ So we can write: SQ = (√3)×RS = (√3)x
8. Now we can consider the original ΔPQR. We get:
[PQ : QR : PR] = [(1+√3)x : 2x : (√2)x]
In this section we have completed the discussion on the trigonometric ratios of special angles 30o, 60o and 45o angles. In the next two sections we will see some solved examples based on this discussion. After those solved examples we will see the discussion on two more special angles 0o and 90o. It can be seen here.
In the fig(a) below, we have a triangle whose angles are 45o, 30o and 105o.
1. Let us name the given triangle as PQR
2. Drop a perpendicular RS from the vertex R on to the side PQ. This is shown in fig(b)
3. Now we have two right triangles: ⊿PSR and ⊿QSR
4. Consider ⊿PSR. We get ∠PRS = 45o.
Consider ⊿QSR. We get ∠QRS = 60o
5. So PSR is a 45o isosceles triangle
QSR is a 30o, 60o triangle
6. Consider ⊿PSR again. Let RS be 'x' cm.
• In a 45o isosceles triangle, we have: Hypotenuse = (√2) times the smallest side. In ⊿PSR, RS is one of the smallest sides because, it is opposite the smallest angle 45o
♦ So we can write: PR = (√2)×RS = (√2)x
Also in a 45o isosceles triangle, the smallest sides are equal.
♦ So we can write: PS = RS = x
These are shown in fig(c)
7. Consider ⊿QSR again
In a 30o, 60o triangle, we have: Hypotenuse = 2 times the smallest side. In ⊿QSR, RS is the smallest side because, it is opposite the smallest angle 30o
♦ So we can write: QR = 2×RS = 2x
Also in a 30o, 60o triangle, the medium side is (√3) times the smallest side. In ⊿QSR. SQ is the medium side because, it is opposite the medium angle 60o
♦ So we can write: SQ = (√3)×RS = (√3)x
8. Now we can consider the original ΔPQR. We get:
[PQ : QR : PR] = [(1+√3)x : 2x : (√2)x]
In this section we have completed the discussion on the trigonometric ratios of special angles 30o, 60o and 45o angles. In the next two sections we will see some solved examples based on this discussion. After those solved examples we will see the discussion on two more special angles 0o and 90o. It can be seen here.
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