Chapter 34 - Geometry and Algebra

In the previous section we completed a discussion on surface area and volume of some solids. In this chapter we will see Geometry and Algebra.

• This chapter is in fact a continuation of chapter 31 - Coordinate Geometry. 
• The last section of that chapter can be seen here. In that chapter 31, we saw an interesting case:
1. We were given the coordinates of two points
• Those were the end points of a diagonal of a rectangle or a square
2. We were asked to find the coordinates of the other two corners of the rectangle
• We solved such problems very easily. (Details here)

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• Now we will see a similar case. 
    ♦ This time, instead of rectangles or squares, we look at parallelograms. 
• Coordinates of any three of the four vertices of a parallelogram will be given. 
• We will be asked to find the coordinates of the fourth vertex. 
Let us see how it is done:
1. Consider the parallelogram OABC in fig.34.1(a) below

Fig.34.1

• Coordinates of three vertices O, A and C are known. We have to find the coordinates of the vertex B
2. In fig.b, green lines are drawn. 
• These green lines have a special property: They are all parallel to the axes
    ♦ The horizontal green lines are parallel to the x axis
    ♦ The vertical green lines are parallel to the y axis
3. Consider the two green lines attached to the side CB
• They intersect at B'. 
• We know that the angle between the axes will always be 90^o
• Since the green lines are parallel to the axes, the angle at B' will also be 90^o
• So triangle BB'C is a right triangle
4. In the same way, the bottom triangle AA'O is also a right triangle
5. Now we will see the relation between the two right triangles:
• OA and CB are the opposite sides of a parallelogram. 
    ♦ So OA and CB are parallel to each other 
    ♦ and they will be inclined at the same angle with the x axis
• CB' and OA' are both parallel to the x axis. So we get: ∠BCB' = ∠AOA'
• These two equal angles are denoted as α in fig.c
6. Again, OA and CB are parallel to each other and they will be inclined at the same angle with the y axis also
• BB' and AA' are both parallel to the y axis. So we get: ∠CBB' = ∠OAA'
• These two equal angles are denoted as β in fig.c
7. We have a situation:
(i) Angles at the ends of the hypotenuse BC are α and β
(ii) Angles at the ends of the hypotenuse OA are also α and β
(iii) The two triangles are right angled. 
(iv) The two hypotenuses, being opposite sides of a parallelogram, have the same lengths
■ In this situation, we can apply the RHS criterion. The two triangles are equal. (Details here)
Let us write the correspondence:
• The 90^o angle is at A' and B'. So A'↔B'
• The α^o angle is at O and C. So O↔C
• The β^o angle is at A and B. So A↔B
• So the correspondence is A'↔B', O↔C and A↔B. This is same as A'OA↔B'CB
• So we can write: 
The two triangles are congruent to one another under the correspondence: A'OA↔B'CB
• This is same as ΔA'OA ≅ ΔB'CB
Now we can write the corresponding sides. We get: 
    ♦ A'O↔B'C
    ♦ OA↔CB
    ♦ A'A↔B'B
8. Among the above three correspondences, the middle one is already known to us. 
• Because, they are the opposite sides of a parallelogram. 
• The first and the last are of great use to us.
■ The first one indicates this:
The bases of the two triangles are of equal length
■ The last one indicates this:
The altitudes of the two triangles are of equal length
9. Now consider fig.34.2(a) below:

Fig.34.2

• AA' is parallel to the y axis. So coordinates of A' are (6,0)
• Thus length of OA' is 6 units. This is shown in fig.34.2(b)
10. We have seen that the bases are equal. 
• So CB' must also be 6 units in length
• Then coordinates of B' will be (3+6, 5). That is., (9,5)
11. Now consider the coordinates of A. They are (6,2)
• So AA' will be 2 units in length
12. We have seen that the altitudes are equal
• So BB' will also be 2 units in length
13. We already wrote the coordinates of B' as (9,5). 
• So the coordinates of B will be (9,5+2). That is., (9,7)
• This is shown in fig.c. Thus we determined the unknown coordinates at B. The problem is solved.

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• We were able to solve it because, the two bases as well as the two altitudes are equal. We proved it using the RHS criterion.
• There is an easier method to prove those equalities. In that method we use trigonometric ratios. We have already seen the basics of trigonometry here.
1. In fig.34.2(a) above, consider the right triangle BCB'. Taking trigonometric ratios, we will get:
sin α = ^{opposite side}⁄_hypotenuse = ^BB'⁄_CB.
cos α = ^{adjacent side}⁄_hypotenuse = ^B'C⁄_CB.
2. Again in fig.34.2(a) above, consider the right triangle AOA'. Taking trigonometric ratios, we will get:
sin α = ^{opposite side}⁄_hypotenuse = ^AA'⁄_OA.
cos α = ^{adjacent side}⁄_hypotenuse = ^A'O⁄_OA.
3. Now, the sine in ⊿BCB' can be equated to the sine in ⊿AOA'. Because, both are taken for the same angle α.
• For example, suppose α be 36^o. Then from the tables, we have: sin 36 = 0.5878
• So in ⊿BCB', sin α = sin 36 = ^BB'⁄_CB = 0.5878
• In ⊿AOA', ^AA'⁄_OA will also be equal to 0.5878. Because:
sin α = sin 36 = ^AA'⁄_OA= 0.5878
4. So we can write: ^BB'⁄_CB = ^AA'⁄_OA.
• But CB = OA. So we get: BB' = AA'
■ That is., altitudes are equal
5. The same is applicable to cosine also:
• Equating the cosines, we get:
^B'C⁄_CB = ^A'O⁄_OA.
• But CB = OA. So we get: B'C = A'O
■ That is., bases are equal

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• So now we are in a position to find the unknown vertex of any parallelogram. 
• The method can be applied very quickly just by doing some metal calculations. 
Let us analyse:
1. In fig.34.2(b), imagine a person standing at O
• To reach A, he must first travel 6 units horizontally to the right
    ♦ This '6 units' is determined from the x coordinate of A 
• Then he must travel 2 units vertically upwards
    ♦ This '2 units' is determined from the y  coordinate of A 
2. Another person at C must follow the same procedure to reach B. That is:
• He must first travel 6 units horizontally to the right
• Then he must travel 2 units vertically upwards
■ The direction of each travel must be clearly specified:
• Horizontal - Towards left OR Towards right
• Vertical - Upwards OR Downwards
3. We are able to use this easy method because:
    ♦ bases are equal
    ♦ altitudes are also equal
• This method will be more clear when we see a solved example

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Solved example 34.1:
What are the coordinates of the vertex C of the parallelogram shown in the fig.34.3 below:

Fig.34.3

Solution:
1. Group the vertices into two: [A,B] and [C,D]
• Coordinates of both vertices in the first group are know. So we can write the details of the travel within that group:
2. To reach B from A:
• First travel 4 units horizontally to the right [∵ (5-1) = 4]
• Then travel 1 unit vertically upwards [∵ (4-3) = 1]
3. The same procedure of travel must be followed for the travel from D to C
• First travel 4 units horizontally to the right
    ♦ At the end of this travel, the coordinates will be (6,5) [∵ (2+4) = 6]
• Then travel 1 unit vertically upwards
    ♦ At the end of this final lap, the coordinates will be (6,6) [∵ (5+1) = 6]
So the coordinates of C are (6,6)

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■ In the above steps, we chose [A,B] and [C,D]
Let us choose [A,D] and [B,C]
• Coordinates of both vertices in the first group are know. So we can write the details of the travel within that group:
2. To reach A from D:
• First travel 1 unit horizontally to the right [∵ (2-1) = 1]
• Then travel 2 units vertically upwards [∵ (5-3) = 2]
3. The same procedure of travel must be followed for the travel from B to C
• First travel 1 unit horizontally to the right
    ♦ At the end of this travel, the coordinates will be (6,4) [∵ (5+1) = 6]
• Then travel 2 units vertically upwards
    ♦ At the end of this final lap, the coordinates will be (6,6) [∵ (4+2) = 6]
So the coordinates of C are (6,6)
• We get the same answer. So we can choose groups in two different ways
■ But it is important to note that, the members of a group should not be diagonally opposite. Then the method will not work

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In the next section, we will see a few more solved examples.

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Trigonometric ratios of 0 and 90 degree Angles

In the previous section we saw solved examples on right triangles with special angles: 30^o, 60^o and 45^o. In this section, we will see two more special angles: 0^o and 90^o.

Consider the circle in fig.1 below.
1. It's center 'O' is at the intersection of two axes: the X axis and the Y axis. These two axes are perpendicular to each other.

Fig.1

2. Consider the radius OB₁.
• It makes an angle θ₁ with the horizontal axis.
3. From B₁, a perpendicular B₁A₁ is dropped onto the horizontal axis.
• So we get a right triangle:  ⊿OA₁B₁
4. Now, if we increase the value of θ, with O as the pivot point, the radius OB₁ will rotate about O.
• Let the new value of θ be θ₂.
• B₁ will move to B₂ and A₁ will move to A₂.
• Here also we get a right triangle: ⊿OA₂B₂
5. Let us take trigonometric ratios in both the triangles:
• In ⊿OA₁B₁, sin θ₁ = ^A₁B₁⁄_OB1
• In ⊿OA₂B₂, sin θ₂ = ^A₂B₂⁄_OB2
6. But denominators are the same because OB₁ = OB₂ (since radii of the same circle)
• Also A₁B₁ < A₂B₂ (since A₂B₂ is nearer to the Y axis)
7. Since denominators are the same, from (5), we can easily see that sin θ₁ is less than sin θ₂
■ In general we can say that, when the angle increases, the sine value increases

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Now we will see the extreme values.
• We have seen that when the angle increases, the sine value increases. Then what is the maximum value?
To find the answer, consider the fig.2 below:

Fig.2

1. The angle θ is now increased to θ₃. This θ₃ is very nearly equal to 90^o.
2. Consider the right triangle ⊿OA₃B₃
• We can see that the opposite side A₃B₃ is very nearly equal to the hypotenuse OB₃
• When θ₃ is exactly equal to 90, A₃B₃ will become exactly equal to OB₃
• So when θ₃ is very close to 90, we can consider A₃B₃ to be very close to OB₃.
    ♦ In such a condition, the difference in lengths between A₃B₃ and OB₃ will be very small like 0.000001 cm
    ♦ Even when the difference in lengths is such a low value, the ⊿OA₃B₃ will exist
   ♦ But if the difference become zero, the ⊿OA₃B₃ will disappear because, A₃B₃ will coincide with OB₃.
•For practical purposes, when the difference between A₃B₃ and OB₃ is very small, we can take:
θ₃ = 90^o and A₃B₃ = OB₃.
3. we will get:
sin θ₃ = sin 90 = ^A₃B₃⁄_OB3 = 1 (since numerator = denominator)
■ That is., we can take sin 90 = 1

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Now we will see the other extreme:

• We have seen that when the angle increases, the sine value increases. It increases upto a maximum of 1. 
• That means, when angle decreases, the sine value decreases. 
• Then what is the minimum value?
To find the answer, consider the fig.3 below:

Fig.3

1. θ is now decreased to θ₀. This θ₀ is very nearly equal to 0^o
2. Consider the right triangle: ⊿OA₀B₀
• We can see that the opposite side A₀B₀ is now very small
• When θ₀ is exactly equal to 0, the length of the opposite side A₀B₀ will become exactly equal to 0 cm
• So when θ₀ is very close to 0, we can consider A₀B₀ to be very close to 0 cm
    ♦ In such a condition, the length of A₀B₀ will be very small like 0.000001 cm
    ♦ Even when the length is such a low value, the ⊿OA₀B₀ will exist
   ♦ But if the length become zero, the ⊿OA₀B₀ will disappear because, A₀B₀ will coincide with OB₀.
•For practical purposes, when the length of A₀B₀ is very small, we can take:
θ₀ = 0^o and A₀B₀ = 0 cm
3. we will get:
sin θ₀ = sin 0 = ^A₀B₀⁄_OB0 = 0 (since numerator = 0)
■ That is., we can take sin 0 = 0

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So we got three information:

• When the angle increases, sine value increases
• The minimum value is: sin 0 = 0
• The maximum value is: sin 90 = 1

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Next we will consider cosine. The same three figs. can be used for this purpose also

1. Consider fig.1 above. For the angle θ, the opposite sides are A₁B₁, A₂B₂ … so on
• As θ increases, these opposite sides increases
2. But the adjacent sides are OA₁, OA₂ . . . so on
• As θ increases, these adjacent sides decreases.
3. Let us see the effect of this decrease when we take the cosine ratios:
• In ⊿OA₁B₁, cos θ₁ = ^OA₁⁄_OB1
• In ⊿OA₂B₂, cos θ₂ = ^OA₂⁄_OB2
4. But denominators are the same because OB₁ = OB₂ (since radii of the same circle)
• Also OA₂ < OA₁ (since A₂ is nearer to the origin than A₁)
5. Since denominators are the same, from (3), we can easily see that cos θ₂ is less than cos θ₁
■ In general, we can say that, when the angle increases, the cosine value decreases

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Now we will see the extreme values for cosine.

• We have seen that when the angle increases, the cosine value decreases. Then what is the minimum value?
To find the answer, consider the fig 2 above
1. The angle θ is now increased to θ₃. This θ₃ is very nearly equal to 90^o.
2. Consider the right triangle ⊿OA₃B₃
• We can see that the adjacent side OA₃ is now very small
• When θ₃ is exactly equal to 90, the length of the adjacent side OA₃ will become exactly equal to 0 cm
• So when θ₃ is very close to 90, we can consider OA₃ to be very close to 0 cm
    ♦ In such a condition, the length of OA₃ will be very small like 0.000001 cm
    ♦ Even when the length is such a low value, the ⊿OA₃B₃ will exist
   ♦ But if the length become zero, the ⊿OA₃B₃ will disappear because, A₃B₃ will coincide with OB₃.
•For practical purposes, when the length of OA₃ is very small, we can take:
θ₃ = 90^o and OA₃ = 0 cm
3. we will get:
cos θ₃ = cos 90 = ^OA₃⁄_OB3 = 0 (since numerator = 0)
■ That is., we can take cos 90 = 0

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Now we will see the other extreme for cosine:

• We have seen that when the angle increases, the cosine value decreases. It decreases upto a minimum of zero. 
• That means, when angle decreases, the cosine value increases. 
• Then what is the maximum value?
To find the answer, consider the fig.3 above.
1. θ is now decreased to θ₀. This θ₀ is very nearly equal to 0^o 
2. Consider the right triangle: ⊿OA₀B₀
• We can see that the adjacent side OA₀ is very nearly equal to the hypotenuse OB₀
• When θ₃ is exactly equal to 0, OA₀ will become exactly equal to OB₀
• So when θ₃ is very close to 0, we can consider OA₀ to be very close to OB₀.
    ♦ In such a condition, the difference in lengths between OA₀ and OB₀ will be very small like 0.000001 cm
    ♦ Even when the difference in lengths is such a low value, the ⊿OA₀B₀ will exist
   ♦ But if the difference become zero, the ⊿OA₀B₀ will disappear because, OA₀ will coincide with OB₀.
• For practical purposes, when the difference between OA₀ and OB₀ is very small, we can take:
θ₃ = 0^o and OA₀ = OB₀.
3. we will get:
cos θ₃ = cos 0 = ^OA₀⁄_OB0 = 1 (since numerator = denominator)
■ That is., we can take cos 0 = 1

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So we got three information:

• When the angle increases, cosine value decreases
• The minimum value is: cos 90 = 0
• The maximum value is: cos 0 = 1

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Next we will consider tangent ratio. This does not need lengthy calculations because we know that:
tan θ = ^{sin θ}⁄_{cos θ}
So we get:
• tan 0 = ^{sin 0}⁄_{cos 0} = ⁰⁄₁ = 0
• tan 90 = ^{sin 90}⁄_{cos 90} = ¹⁄₀ . So tan 90 is not defined because it is a division by zero.

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So we have completed this present discussion on the two special angles 0^o and 90^o.

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Chapter 30.7 - Problems related to Trigonometric ratio Tan

In the previous section we saw trigonometry related to circles. In the discussion so far in this chapter, we have seen problems related to sine and cosine. We have seen the basic details of tangent also. Details here. In this section, we will see  some problems related to tangent.

An example:
• Some steps are built on the side of a hill. See fig.30.33(a) below:

Fig.30.33

• The steps lead up to the top of the hill. But all the steps are not shown in the fig. We need only the first few steps for our present problem.
• A cone (coloured in yellow) is placed on the third step. We want to know the level at which the base of the cone is situated. In other words, what is the value of 'h'?
• For solving this problem, we are given two information:
    ♦ The tread of each step is 20 cm
    ♦ The angle which the side of the hill makes with the horizontal ground surface is 35^o
Solution:
1. Drop a perpendicular  CB from the edge of the third step onto the ground surface. This is shown by the red dashed line in fig.(b)
• Now we have a right triangle: ⊿ABC. 
2. Base AB will be equal to 3 times the tread. So we get AB = 3 × 20 = 60 cm
3. In this right triangle,
tan 35 = ^{opposite side}⁄_{adjacent side}  = ^BC⁄_AB  = ^h⁄₆₀  ⟹ h = 60 × tan 35 ⟹ h = 60 × 0.7002 = 42.01 cm

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Now we will see a few solved examples like this:

Solved example 30.21

An iron rod of unknown length, leans against a wall. The bottom end of the iron rod is 2 m away from the wall. Also it makes an angle of 40^o with the floor. See fig.30.34(a) below:

Fig.30.34

How high is the top end of the iron rod from the ground?
Solution:
1. The wall, iron rod and the ground surface together will give a right triangle as shown in fig.(b)
We have tan 40 = ^{opposite side}⁄_{adjacent side}  = ^h⁄₂  ⟹ h = 2 × tan 40 ⟹ h = 2 × 0.8391 = 1.68 m

Solved example 30.22 
Three rectangles are cut along the diagonals as shown in the figs.30.35(a, b and c) below:

Fig.30.35

One large rectangle and two smaller rectangles. The two smaller rectangles are identical. The 6 pieces so obtained are rearranged to form a regular pentagon. See. fig.(d). If the sides of the pentagon are to be 30 cm, what should be the length and breadth of the rectangles  
Solution:
1. First we want the interior angles of the pentagon. Since it is a regular pentagon, all sides will be equal. All angles will also be equal.
• We have: Sum of interior angles of any regular polygon = (n-2)×180. 
Where n is the number of sides.
• For a regular pentagon, n = 5. So the sum of all the interior angles = (5-2)×180 = 3×180 = 540^o
    ♦ So angle at each vertex = ⁵⁴⁰⁄₅ = 108^o
2. Now consider the yellow and magenta triangles in fig.30.36 below:

Fig.30.36

3. They are identical. So the edge AB will bisect the interior angle (at vertex A) of the pentagon. So we get: ∠CAB = ∠DAB = ¹⁰⁸⁄₂ = 54^o
4. Consider ⊿ABC. 
sin 54 = ^{opposite side}⁄_hypotenuse  = ^BC⁄_AC = ^BC⁄₃₀  ⟹ BC = 30 × sin 54 
⟹ BC = 30 × 0.8090 = 24.27 cm
■ So we get: Length of the small rectangles = 24.27 cm
5. Again consider ⊿ABC. 
cos 54 = ^{adjacent side}⁄_hypotenuse  = ^AB⁄_AC = ^AB⁄₃₀  ⟹ AB = 30 × cos 54 
⟹ AB = 30 × 0.5878 = 17.63 cm
■ So we get: Width of the small rectangles = 17.63 cm
6. Now consider the vertex C of the pentagon. The yellow and green triangles are identical. So each of them will contribute 36^o at the vertex C
7. The blue and red triangles are identical. So they will contribute equally at vertex C
So we can write:
2×36 + 2 × contribution from blue triangle = 108 
⟹ 72 + 2 contribution from blue triangle = 108 
⟹ 2 × contribution from blue triangle = 36 
⟹ contribution from blue triangle = ³⁶⁄₂  = 18^o
• So contribution from red triangle is also 18^o
8. Now consider triangle DEC. It is right angled at E. So we get:
∠CDE + 90 + 18 = 180 ⟹ ∠CDE + 108 = 180 ⟹ ∠CDE = 72^o
9. tan 72 = ^CE⁄_DE
• But DE = half of the side of the pentagon = 15 cm
• So we get: tan 72 = ^CE⁄₁₅ ⟹ CE = 15 × tan 72 ⟹ CE = 15 × 3.078 = 46.17 cm
■ So length of the large rectangle is 46.17 cm and it's width is 15 cm

Solved example 30.23
In the fig.30.37(a) below, the vertical lines are equally spaced.

Fig.30.37

Prove that their heights are in arithmetic sequence
Solution:
1. In fig.(b), the vertical lines are marked as AE, BF, CG etc.,
2. These vertical lines are drawn at an equal spacing of 's' cm
• Let the initial distance OA be 'a' cm
3. Consider the first triangle OAE. Since AE is vertical, triangle OAE is right angled at A
• So we get: tan 40 = ^AE⁄_a ⟹ AE = a tan 40
4. Consider the second triangle OBF. Since BF is vertical, triangle OBF is right angled at B
• So we get: tan 40 = ^BF⁄_(a+s) ⟹ BF = (a+s) tan 40
5. Consider the third triangle OCG. Since CG is vertical, triangle OCG is right angled at C
• So we get: tan 40 = ^CG⁄_(a+s+s) ⟹ BF = (a+s+s) tan 40 ⟹ BF = (a+2s) tan 40
6. Consider the fourth triangle ODH. Since DH is vertical, triangle ODH is right angled at D
• So we get: tan 40 = ^DH⁄_(a+s+s+s) ⟹ DH = (a+s+s+s) tan 40 ⟹ DH = (a+3s) tan 40
7. Continuing like this, if there are more vertical lines, we will get their heights as:
(a+4s) tan 40, (a+5s) tan 40 etc.,
8. Let us write the heights as a sequence:
a tan 40, (a+s) tan 40, (a+2s) tan 40, (a+3s) tan 40 . . . ,
9. In the above sequence, only 4 terms are written. We can write more terms if we want, just by increasing the coefficient of 's'.
• Now we have to prove that this sequence is an arithmetic sequence
10. Fourth term - third term = (a+3s) tan 40 - (a+2s) tan 40
= tan 40[(a+3s)-(a+2s)]
= tan 40[a+3s-a-2s]
= tan 40[s]
= s tan 40
11. Third term - second term = (a+2s) tan 40 - (a+s) tan 40
= tan 40[(a+2s)-(a+s)]
= tan 40[a+2s-a-s]
= tan 40[s]
= s tan 40
12. Second term - first term = (a+s) tan 40 - (a) tan 40
= tan 40[(a+s)-(a)]
= tan 40[a+s-a]
= tan 40[s]
= s tan 40
13. So we find that the difference the difference between any term and it's preceding term is a constant.
• So it is an arithmetic sequence
• The common difference is s tan40.

Solved example 30.24
One side of a triangle is 6 cm and the angles at it's ends are 40 and 65. Calculate it's area
Solution:
1. The given data is shown in fig.30.38(a) below:

Fig.30.38

2. Let us name the triangle as ABC. Drop a perpendicular CD from the vertex C on to the base AB
• Then area of ΔABC = ¹⁄₂ × AB × CD
• So we have to find CD
3. In ΔADC we have:
tan 40 = ^CD⁄_AD ⟹ AD = ^CD⁄_{tan 40}
4. In ΔBDC we have:
tan 65 = ^CD⁄_BD ⟹ BD = ^CD⁄_{tan 65}
5. But AB = AD + BD ⟹ 6 = [^CD⁄_{tan 40} + ^CD⁄_{tan 65}] ⟹ 6 = [^CD⁄_{tan 40} + ^CD⁄_{tan 65}]
⟹ 6 = [^CD⁄_0.8391 + ^CD⁄_2.1446]
[^(CD×2.1446)⁄_{(0.8391× 2.1446)} + ^(CD×0.8391)⁄_{(0.8391×2.1446)}] = 6
⟹  [^(CD×2.1446)⁄_{(0.8391× 2.1446)} + ^(CD×0.8391)⁄_{(0.8391×2.1446)}] = 6
⟹ [(CD×2.1446) + (CD×0.8391)] = 6×0.8391×2.1446
⟹ (2.9837 × CD) = 10.7972
⟹ CD = 3.62 cm
6. So area = ¹⁄₂ × AB × CD = ¹⁄₂ × 6 × 3.62 = 10.86 cm²

Solved example 30.25
In the fig.30.39(a) below, prove that h = √(xy)

Fig.30.39

Solution:
1. The given triangle is named as ABC in fig.(b). Drop a perpendicular CD from the vertex C onto the base AB
2. ADC is a right triangle.
We have: ∠CAD + 90 + ∠ACD = 180 ⟹ 30 + 90 + ∠ACD = 180^o ⟹ ∠ACD = 60^o
3. So it is a 30, 60, 90 triangle
• CD is the shortest side (opposite to the smallest angle 30)
• AD is the medium side (opposite to the medium angle 60)
• AC is the largest side (opposite to the largest angle 90)
4. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• AD = √3 × CD ⟹ x = √3 × h = h√3 cm
5. BDC is a right triangle.
We have: ∠CBD + 90 + ∠BCD = 180 ⟹ 60 + 90 + ∠BCD = 180^o ⟹ ∠BCD = 30^o
6. So it is a 30, 60, 90 triangle
• BD is the shortest side (opposite to the smallest angle 30)
• CD is the medium side (opposite to the medium angle 60)
• BC is the largest side (opposite to the largest angle 90)
7. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• CD = √3 × BD ⟹ h = √3 × y = y√3 ⟹ y = ^h⁄_(√3)
8. From (4) we have: x = h√3
From (7) we have: y = ^h⁄_(√3)
9. So xy = h√3 × [^h⁄_(√3)] = h² ⟹ h = √(xy)

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In the next section we will see another application of tangent.

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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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