Find the value of X in each of the following instances:
900
X
900
20
5
900
10
24
X
900
3
12
900
900
X
Solution 1.
Consider the large triangle below. Let the vertices be A, B and C. Let AD be the line that is
drawn perpendicular to BC.
A
Y
900
Z
X
900
B
20
D
5
C
Consider the larger right angled triangle ABC. Let BA = Y and AC = Z
Since the triangle ABC is a right angled triangle (Angle A = 900)Using Pythagoras Theorem
BC2 = AB2+AC2
BC = BD+DC = 20+5 = 25
252 = Y2+Z2 --------------------------------------- 1
Consider the triangle ABD. It is also a right angled triangle as angle ADB = 900
Using Pythagoras Theorem we can write
AB2= BD2+AD2
Y2 = 202+X2 -------------------------------------- 2
Consider the triangle ADC. It is also a right angled triangle as angle ADC = 900
Using Pythagoras Theorem we can write
AC2= DC2+AD2
Z2 = X2+52 -------------------------------------- 3
Add the equations 2 and 3. Adding equation 2 and 3 will give the value of Y2 + Z2 in terms of X.
Adding
Y2 + Z2 = (202+X2)+( X2+52)
We know from equation 1 that 252 = Y2+Z2. Substituting Y2 + Z2 = 252 in the above equation we
get
252 = (202+X2)+( X2+52) or 252 - 202 - 52 = 2 (X2)
652-400-25 =2 (X2) or 200 = 2 (X2) or X2= 100 or X = 10 units
Solution 2:
Consider the large triangle below. Let the vertices be A, B and C. Let AD be the line that is
drawn perpendicular to BC.
A
900
10
24
X
900
a
B
b
D
C
Y
Consider the triangle ABC. Let BC = Y length of the side opposite to the 900 angle
Using Pythagoras Theorem BC2 = AB2+AC2
Y2 = 102+242 or Y2 = 100+576 or Y2 = 676 or Y = 26 units ---------------------------------- 1
Consider the smaller right angled triangles ABD and ADC. Let BD = a and DC = b. From the
above figure
BD+DC = BC or a+b = Y or a+b = 26 (since Y = 26 from equation 1) ----------------------------- 2
Consider the triangle ABD. The side opposite to the 900 is AB. Using Pythagoras Theorem
AB2 = BD2+AD2 or 102 = a2+X2 (since BD = a and AD =X) or X2 = 102-a2 -------------------------- 3
Consider the triangle ADC. The side opposite to the 900 is AC. Using Pythagoras Theorem
AC2 = DC2+AD2 or 242 = b2+X2
(since DC = b and AD =X) or X2 = 242-b2 ------------------------ 4
Equation 3 and 4 should be equal. So we can write
102-a2 = 242-b2 or b2 – a2 = (242-102) or b2-a2 = 476 ------------------------------ 5
Using the algebraic identity b2-a2 = (b-a)(b+a) we can write equation 5 as
(b-a)(b+a) = 476. From equation 2 we know that a+b = 26. Substituting a+b = 26 we get
(b-a) x 26 = 476 or b-a = 476/26 or b-a = 18.3076 ------------------------ 6
Solving for a and b using the equations 2 and 6
b+a =26
b-a = 18.3076
Adding these two equations we get
2b = 44.3076 or b = 22.1538 ---------------------- 7
Substituting the value of b in equation 4 i.e. X2 = 242-b2
X2 = 242- (22.1538)2 = 85.2071 or X = 9.23 units
Alternative Solution – Using Trigonometric Function
A
900
10
24
X
900
θ
a
B
b
D
C
Y
Consider the triangle ABC. Let θ be the angle ABC. Since ABC is a right angled triangle and BC
is the hypotenuse of the triangle.
Tan θ = Opp/Adj where
Opp = Side Opposite to θ = AC = 24 units
Adj = side adjacent to θ = AB = 10 units
So Tan θ = AC/AB = 24/10 = 2.4 or Tan θ = 2.4
In order to find the value of θ, we need to use the tan-1 function.
Tan-1 function is such that if Tan θ = a than tan-1 a = θ.
We can write if Tan θ = 2.4 than tan-1(2.4) = θ or θ = 67.380
Now consider the triangle ADB. We need to find the value of AD (i.e. X). AD is the side opposite
to angle θ.
We can write sin θ = Opp/Hyp
Opp = Side Opposite to θ = AD= X units
Hyp = Hypotenuse = AB = 10 units
θ = 67.380
So sin 67.380 = Opp/Hyp = X/10
Using the calculator Sin 67.38 = 0.9230
So 0.9230 = X/ 10 or X = 9.23 units which is the same as that obtained using the
Pythagoras method
Solution 3: Consider the large triangle below. Let the vertices be A, B and C. Let AD be the line
that is drawn perpendicular to BC.
C
3
D
12
900
a
B
Y
900
X
A
Consider the triangle ABC. Let AC = Y units. Also BC = Side opposite to the 900 = BD+DC
BC = 12+3 = 15 units
Using Pythagoras Theorem BC2 = AB2+AC2
152 = X2+Y2 ----------------------------------------------- 1
AD is drawn perpendicular to BC and let length of AD = a units. So the Triangle ABC can be
divided into two right angled triangles ABD and ACD
Consider the triangle ABD. AB = hypotenuse = X units, AD = a units and BD = 12 units
Using the Pythagoras Theorem AB2 = AD2+BD2 or X2 = a2+122 --------------------------- 2
Consider the triangle ACD. AC = hypotenuse = Y units, AD = a units and CD = 3 units
Using the Pythagoras Theorem AC2 = AD2+CD2 or Y2 = a2+32 ------------------------------ 3
Adding equation 2 and 3 we get
X2+Y2 =(a2+122) + (a2+32)
We know from equation 1 X2+Y2 = 152. So we can write the above equation as
152 =(a2+122) + (a2+32) or 152 - 122 - 32 = 2(a2) or 2(a2) = 72 or a2 = 36 or a = 6 units --------- 4
Substituting the value of a = 6 in equation 2 we get
X2 = 62+122 or X2 = 36+144 or X2 = 180 or X = 6√5