Chapter 30.5 - Solved examples on 30, 60 and 45 degree Triangles

In the previous section we saw details about 30^o, 60^o and 45^o right triangles. In this section, we will see some solved examples based on that discussion.

Solved example 30.13
Calculate the areas of the parallelograms shown in fig.30.16 below

Fig.30.16

Solution:
Case 1:
1. Let us name the parallelogram as ABCD. Area of parallelogram ABCD = Base × Height
2. But height is not given. So we drop a perpendicular DE from vertex D onto the side AB. This is shown in fig.30.17(a) below:

Fig.30.17

3. So now we have a right triangle: ⊿AED
4. Consider ⊿AED. For the angle 45^o, opposite side is DE. Hypotenuse of the triangle is AD. So we can write:
sin 45 = ^{opposite side}⁄_hypotenuse = ^DE⁄_AD = ^DE⁄₂
5. But sin 45 = ¹⁄_√2
6. Equating (4) and (5) we get: ^DE⁄₂ = ¹⁄_√2 ⟹ DE = ²⁄_√2 = ^(√2×√2)⁄_√2 = √2 cm
7. So the required area = Base × Height = 4 × √2 = 4√2 cm².

Case 2:
1. Let us name the parallelogram as PQRS. Area of parallelogram PQRS = Base × Height
2. But height is not given. So we drop a perpendicular ST from vertex S onto the side PQ. This is shown in fig.30.17(b) above.
3. So now we have a right triangle: ⊿PTS
4. Consider ⊿PTS. For the angle 60^o, opposite side is ST. Hypotenuse of the triangle is PS. So we can write:
sin 60 = ^{opposite side}⁄_hypotenuse = ^ST⁄_PS = ^ST⁄₂
5. But sin 60 = ^√3⁄₂
6. Equating (4) and (5) we get: ^ST⁄₂ = ^√3⁄₂ ⟹ ST = √3 cm
7. So the required area = Base × Height = 4 × √3 = 4√3 cm².

Solved example 30.14
A rectangular board is to be cut along the diagonal and pieces so formed should be rearranged to form an equilateral triangle. See fig.30.18 below:

Fig.30.18

Sides of the triangle must be 50 cm. What should be the length and width of the original rectangle in fig(a)? 
Solution:
1. Let us name the rectangle as PQRS. See fig.30.19(a) below:

Fig.30.19

The rectangle is cut along the diagonal PR
2. The ⊿PQR is kept stationary. The other ⊿PSR is shifted and placed in such a way that the top edge SR becomes aligned with PQ. Now we get a triangle.
3. But this triangle must be equilateral. We know that angles in an equilateral triangle are 60^o.
4. So, the left right triangle, the angle at P will be 60^o.
5. Also, the sides must be 50 cm. So diagonal of the original rectangle must be 50 cm.
6. Blue and red edges must be 25 cm each. That means, width of the original rectangle must be 25 cm
5. From the left side right triangle we get: sin 60 = ^{opposite side}⁄_hypotenuse 
= ^{Green edge}⁄₅₀ 
6. But sin 60 = ^√3⁄₂.
7. Equating (5) and (6) we get: ^{Green edge}⁄₅₀ = ^√3⁄₂ ⟹ Green edge = 25√3 cm
8. So the length of the original rectangle must be 25√3 cm and it's width must be 25 cm

Solved example 30.15
Two identical rectangles are cut along the diagonal and the pieces so formed are joined to another rectangle to make a regular hexagon. See fig.30.20 below:

Fig.30.20

Sides of the regular hexagon must be 30 cm. What should be the length and width of the rectangles?
Solution:
1. The various pieces are shown in fig.30.21(a) below:

Fig.30.21

2. Fig.30.21(b) shows the method of forming the regular hexagon. It is clear that, the width of the original larger (purple coloured) rectangle must be 30 cm
3. Now we want angles. Sum of interior angles of a regular polygon = (n-2)×180. Where n is the number of sides
• So for a regular hexagon, sum of interior angles = (6-2)×180 = 4×180 = 720
• So angle at each vertex = ^{sum of interior angles}⁄_{number of vertices} = ⁷²⁰⁄₆ = 120^o.
4. Consider the green and blue triangles in fig(c). They are right triangles. Their bases bisect the 120^o into 60^o and 60^o.
5. Consider the green triangle alone. In this triangle, sin 60 = ^{opposite side}⁄₃₀ 
6. But sin 60 = ^√3⁄₂.
7. Equating (5) and (6) we get: ^{opposite side}⁄₃₀ = ^√3⁄₂ ⟹ opposite side = 15√3 cm
8. But this opposite side is the length of the original small rectangles. So we get:
• Length of the original small rectangles = 15√3
9. Also, this opposite side is half the length of the original bigger rectangle. So we get:
• Length of the original bigger rectangle = 30√3
10. Again consider the green triangle alone. In this triangle, cos 60 = ^{adjacent side}⁄₃₀ =
6. But cos 60 = ¹⁄₂
7. Equating (5) and (6) we get: ^{adjacent side}⁄₃₀ = ¹⁄₂ ⟹ adjacent side = 15 cm
8. But this adjacent side is the width of the original small rectangles. So we get:
• Width of the original small rectangles = 15 cm
9. Let us write all the required lengths together:
■ Small rectangles:
Length = 15√3 cm
width = 15 cm
■ Large rectangle:
Length = 30√3 cm
Width = 30 cm

Solved example 30.15
Calculate the area of the triangle shown in fig.30.22(a) below:

Fig.30.22

Solution:
• Let us name the triangle as PQR
1. Area of ΔPQR = ¹⁄₂ × Base × Altitude
2. But altitude is not given. So we drop a perpendicular RS from vertex R onto the side PQ. This is shown in fig(b)
3. So now we have two right triangles: ⊿PSR and ⊿QSR. Let PS = x cm. Then QS = (4-x) cm
4. Consider ⊿PSR. For the angle 45^o, opposite side is RS. Adjacent side is PS. So we can write:
tan 45 = ^{opposite side}⁄_{adjacent side} = ^RS⁄_PS = ^RS⁄_x.
5. But tan 45 = 1
6. Equating (4) and (5) we get: ^RS⁄_x = 1 ⟹ RS = x cm
7. Consider ⊿QSR. For the angle 60^o, opposite side is RS. Adjacent side is QS. So we can write:
tan 60 = ^{opposite side}⁄_{adjacent side} = ^RS⁄_QS = ^RS⁄_(4-X)
8. But tan 60 = √3
9. Equating (7) and (8) we get: ^RS⁄_(4-x) = √3 ⟹ RS = (√3)(4-x) cm
10. Equating (6) and (9) we get: x = (√3)(4-x) ⟹ x = 4√3 - x√3 ⟹ x + x√3 = 4√3
⟹ x(1+√3) = 4√3 ⟹ x = ^(4√3)⁄_(1+√3)
11. So altitude RS = x = ^(4√3)⁄_(1+√3) 
12. So the required area = ¹⁄₂ × Base × Altitude = ¹⁄₂ × 4 × [^(4√3)⁄_(1+√3)] = [^(8√3)⁄_(1+√3)] cm².

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More solved examples on this topic can be seen here.

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In the next section we will see how the trigonometric ratios can be applied to circles.

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Chapter 30.4 - Trigonometric ratios of 30, 60 and 45 degree Angles

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 30^o, 60^o and 45^o.

• First we will consider right triangle with 30^o. When one of the acute angles in a right triangle is 30^o, the other acute angle will be obviously 60^o. 
• This is very convenient for us. We can discuss 30^o and 60^o together.
Fig. 30.13(a) below shows a ⊿ABC. It is right angled at B. The angle at vertex C is 30^o. So the angle at vertex A will be 60^o.

Fig.30.13

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) = 60^o
• Since D is the mirror image of A, Angle at vertex D = 60^o
• Thus in ADC, all the three angles are 60^o. 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

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So now we know the trigonometric ratios of a 30^o, 60^o 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

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• Now we will consider right triangle with 45^o. When one of the acute angles in a right triangle is 45^o, the other acute angle will be obviously 45^o.
• Fig.30.14 shows a ⊿ABC with acute angles 45^o.

Fig.30.14

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

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In the case of 45^o 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.

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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 30^o, 60^o 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 30^o is the smallest angle in a 30^o, 60^o triangle. So the side opposite 30^o will be the smallest side in a 30^o, 60^o triangle. Thus we can write:
■ In a 30^o, 60^o triangle, the hypotenuse is always twice the smallest side

Another result:
1. Consider any 30^o, 60^o 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 30^o is the smallest angle in a 30^o, 60^o triangle. So the side opposite 30^o will be the smallest side in a 30^o, 60^o triangle. 
• Also, the side adjacent to 30 will be the medium side in any 30^o, 60^o triangle. Thus we can write:
■ In a 30^o, 60^o 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 30^o, 60^o triangle, Then:
    ♦ hypotenuse = 2x
    ♦ medium side = (√3)x
• In other words, the ratio smallest side : medium side : hypotenuse in a 30^o, 60^o triangle is:
x : (√3)x : 2x 
⟹ The ratio [smallest side : medium side : hypotenuse] in a 30^o, 60^o triangle is: 
[1 : √3 : 2]
[Note that 2 > √3. So hypotenuse is indeed the longest side in a 30^o, 60^o triangle. In fact, the hypotenuse is the longest side in any right triangle. Because, it is opposite the largest angle (90^o) in a right triangle]

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Next we will derive a similar result for the 45^o isosceles triangle  

1. Consider any 45^o 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 45^o is the smallest angle in a 45^o isosceles triangle. So the side opposite 45^o will be the smallest side in a 45^o isosceles triangle. Thus we can write:
■ In a 45^o isosceles triangle, the hypotenuse is always √2 times the smallest side
• But in a 45^o 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 45^o isosceles triangle, Then:
    ♦ other smallest side = x
    ♦ hypotenuse = (√2)x
• In other words, the ratio smallest side : medium side : hypotenuse in a 45^o isosceles triangle is:
x : x : (√2)x 
⟹ The ratio [smallest side : medium side : hypotenuse] in a 45^o isosceles triangle is: 
[1 : 1 : √2]
[Note that √2 > 1. So hypotenuse is indeed the longest side in a 45^o isosceles triangle]

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In the above discussion, we used two names frequently:

• 30^o, 60^o triangle
• 45^o isosceles triangle
■ On seeing the name 30^o, 60^o triangle we can immediately understand that it is a right triangle
■ On seeing the name 45^o isosceles triangle we can immediately understand that it is a right triangle
The reader may write the reasons in his/her own notebooks

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Now we will see an example:
In the fig(a) below, we have a triangle whose angles are 45^o, 30^o and 105^o.

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 = 45^o.
Consider ⊿QSR. We get ∠QRS = 60^o
5. So PSR is a 45^o isosceles triangle 
QSR is a 30^o, 60^o triangle
6. Consider ⊿PSR again. Let RS be 'x' cm.
• In a 45^o 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 45^o 
    ♦ So we can write: PR = (√2)×RS = (√2)x
Also in a 45^o 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 30^o, 60^o triangle, we have: Hypotenuse = 2 times the smallest side. In ⊿QSR, RS is the smallest side because, it is opposite the smallest angle 30^o
    ♦ So we can write: QR = 2×RS = 2x
Also in a 30^o, 60^o triangle, the medium side is (√3) times the smallest side. In ⊿QSR. SQ is the medium side because, it is opposite the medium angle 60^o
    ♦ 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]

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In this section we have completed the discussion on the trigonometric ratios of special angles 30^o, 60^o and 45^o 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 0^o and 90^o. It can be seen here.  

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Chapter 30.3 - Basic trigonometry - Solved examples

In the previous section we completed a discussion on tangent of an acute angle. We also saw some solved examples. In this section we will see a few more solved examples based on the discussions that we have had so far in this chapter.

Solved example 30.7
What is the area of the ΔABC shown in fig.30.10(a) below:

Fig.30.10

Solution:
• In this problem, it is not known whether ΔABC is right angled or not
1. Area of ΔABC = ¹⁄₂ × Base × Altitude
2. But altitude is not given. So we drop a perpendicular AD from vertex A onto the side BC. This is shown in fig(b)
3. So now we have two right triangles: ⊿ABD and ⊿ACD
4. Consider ⊿ABD. For the angle 50^o, opposite side is AD. Hypotenuse of the triangle is AB. So we can write:
sin 50 = ^{opposite side}⁄_hypotenuse = ^AD⁄_AB = ^AD⁄₄.
5. But from the table, sin 50 = 0.7660
6. Equating (4) and (5) we get: ^AD⁄₄ = 0.7660 ⟹ AD = 0.7660 × 4 = 3.064 cm
7. So the required area = ¹⁄₂ × Base × Altitude = ¹⁄₂ × 6 × 3.064 = 9.192 cm²

Solved example 30.8
If in the above problem, the angle at A is 130^o, what would be the area?
Solution:
If the angle at A is 130, the ΔABC would be as shown in fig.30.11(a) below:

Fig.30.11

1. In this problem also, area of ΔABC = ¹⁄₂ × Base × Altitude
2. But altitude is not given. So we drop a perpendicular AD from vertex A onto the side BC. For that, we extend BC towards the left This is shown in fig(b)
3. So now we have two triangles: ⊿ABD and ΔABC
4. Consider ⊿ABD.The angles ∠ABD and ∠ABC are co-linear. That is., ∠ABD + ∠ABC = 180^o
So we get: ∠ABD = 180 - ∠ABC = 180 - 130 = 50^o
5. For this angle 50^o, opposite side is AD. Hypotenuse of the triangle is AB. So we can write:
sin 50 = ^{opposite side}⁄_hypotenuse = ^AD⁄_AB = ^AD⁄₄.
5. But from the table, sin 50 = 0.7660
6. Equating (4) and (5) we get: ^AD⁄₄ = 0.7660 ⟹ AD = 0.7660 × 4 = 3.064 cm
7. So the required area = ¹⁄₂ × Base × Altitude = ¹⁄₂ × 6 × 3.064 = 9.192 cm²

Solved example 30.9
Find the length of the side QR of the ΔPQR in fig.30.12(a) below:

Fig.30.12

Solution:
• In this problem, it is not known whether ΔPQR is right angled or not
1. Drop a perpendicular RS from vertex R onto side PQ. Now we have two right triangles: ⊿PRS and ⊿QRS
2. Consider ⊿PRS. For the angle 40^o, opposite side is RS. Hypotenuse of the triangle is PR. So we can write:
sin 40 = ^{opposite side}⁄_hypotenuse = ^RS⁄_PR = ^RS⁄₆.
3. But from the table, sin 40 = 0.6428
4. Equating (2) and (3) we get: ^RS⁄₆ = 0.6428 ⟹ RS = 0.6428 × 6 = 3.8568 cm
5. Consider ⊿PRS again. For the angle 40^o, adjacent side is PS. Hypotenuse of the triangle is PR. So we can write:
cos 40 = ^{adjacent side}⁄_hypotenuse = ^PS⁄_PR = ^PS⁄₆.
6. But from the table, cos 40 = 0.7660
7. Equating (5) and (6) we get: ^PS⁄₆ = 0.7660 ⟹ PS = 0.7660 × 6 = 4.596 cm 
8. Now we can find SQ:
SQ = PQ - PS = 7 - 4.596 = 2.404 cm
9. Consider ⊿QRS. Applying Pythagoras theorem, we get:
QR² = SQ² + SR² ⟹ QR² = 2.404² + 3.8568² ⟹ QR² = 20.654 ⟹ QR = √20.654 = 4.544 cm

Solved example 30.10
This is given in the form of a video presentation. It can be seen here

Solved example 30.11
This is given in the form of a video presentation. It can be seen here 

Solved example 30.12
This is given in the form of a video presentation. It can be seen here  

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In the next section we will see details of triangles with 30^o, 60^o and 45^o.

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