Classification of Triangles According to Their Angles For the classification of triangles as acute triangle, right triangle or obtuse triangle, we make use of Pythagorean Theorem. 1. Acute Triangle
Mark the longest side. In figure 7-61, the longest side is 6. The other two sides are 4 and 5. Now, find the square of the longest side. Here it is 62 which is equal to 36. Find the sum of the squares of the other two sides. Here it is 42 + 52 = 16 + 25 25 = 41 41
We find that 36<41, ∴ ABC in figure 7-61 is an acute triangle. So, we conclude. If in a triangle, the square of the longest side is less than the sum of the squares of the two remaining sides, then the triangle is acute triangle. 2. Right Triangle
Mark the longest side in the triangle. Find its square. In figure 7-62, the longest side is 5, Therefore its square is 52 which is equal to 25. Find the sum of the 2 5 . Now we find squares of the two remaining sides, Here it is 32 + 42 = 9 + 16 = 25 that 25=25. Therefore Triangle ABC is a right triangle. So, we conclude that if in a triangle, the square of the longest side is equal to the sum of the square of the two remaining sides, and then the triangle is a right triangle.
3. Obtuse Triangle
Mark the longest side. Find its square. Find the sum of the squares of the other two sides of the given triangles. Here, the longest side is 6, its square is 62 = 36 . Sum of the squares of the two other sides is 22 + 32 = 4 + 9 = 13 . Now 36>13. Therefore Triangle ABC in figure 7-63 is an obtuse triangle.
So, finally we conclude that If the sides of a triangle are a, b and c . Suppose c is the longest side then if 2
<
a
c2
=
a 2 + b 2 then the triangle is right triangle and if
c2
>
a 2 + b 2 , then the triangle is obtuse triangle.
c
2
+b
2
then the triangle is an acute triangle
Special Right Triangles 1. 30o − 60o − 90o Triangles:
Triangle ABC is In figure 7-64 a right triangle in o o o o which ∠ A= 30 , ∠ B= 60 , ∠ C= 90 . On BC, construct ∠ BCD = 60 . Now o
o
∠ BDC = 180 − (60 +
O
60 )
or ∠ BDC = 180o − 120o = 60O Therefore Triangle DBC is an equilateral triangle
If BC= a, then CD = BD = a Now, in CDA, ∠A = 30o given ∠ DCA =
90
o
o
− 60 =
o
30
Therefore Therefore Triangle CDA is an isosceles triangles
Therefo reforre CD = AD (In an isosceles triangle sides opposite equal angles are
equal)
or CD = a or AD= a
Now AB=AD+DB=a+a Therefore AB=2BC So, we conclude that in a 30o − 60o − 90o triangle, the hypotenuse is twice the sides opposite 30o
From the above figure equilateral Triangle PQR .
it
is
evident
that
Triangle ABC is
Now in Triangle ABC by Pythagoras theorem, we have a
2
2
+b =
c
2
or a 2 + b 2
=
(2 a )
or a 2 + b2
=
4a
or b2
2
=
4a
2
( Since
2
2
−a =
3a
2
c = 2a )
half
the
or b = a 3 Now, if a = 1 unit than c = 2 units and b = 3 units Therefore
a :b : c = 1: 3 : 2
Properties of 30o − 60o − 90o triangle:
1. Side opposite 30o angle equals one-half the length of the hypotenuse a =
1 2c
.
2. Side opposite 60o angle equals one-half the length of the hypotenuse times the square root of 3. b =
3 2
c.
3. The length of the leg opposite the 90o angle equals the length of the side opposite the 30o angle times the square root of 3 b = 3a 4. Length of hypotenuse equals twice the length of side opposite 30o angle. 2. 45o − 45o − 90o triangle:
In Triangle ABC , ∠B = 45o , ∠C = 45o
Therefore Triangle ABC is an isosceles triangle, and in an isosceles triangle side
opposite equal angles are equal. Let BC= a, then AC= a Now, in Triangle ABC 2
2
2
(Pythagoras Theorem)
AC + BC = AB
or a 2 + a 2
=
or 2a 2
2
=c
c2
Length of the hypotenuse = 2 × length of the side opposite 30o = 2 × 5 = 10 cm Now applying property 3, we have Length of the side opposite 60o =
=
o
3 times length of the side opposite 30 angle
3 × 5 = 5 3 cm
How to find the sides of a right triangle when side opposite 60o is given
Here we have to find the other sides in terms of b .
1
Rule1: a = b 3 3
Side opposite 30o angle is one-third of square root of 3 times the side opposite o 60 angle. Rule 2: The hypotenuse is two-thirds of square root of 3 times the side opposite o 60 angle c=
1 3
2 3b
SOLVED EXAMPLES 1. In a right triangles, the side opposite 60o angle is 9 cm. Find the other sides. Sol:
Using Rule 1, we have 1
1
3
3
The side opposite 30o angle = a = b 3 =
× 9×
3 = 3 3 cm
Using Rule 2, we have the hypotenuse = c =
2 3
3b =
2 3
3 × 9 = 6 3 cm
2. In a 30o − 60o − 90o triangle, the side opposite 60o angle is 15 cm. Find the other two sides. Verify you answer Sol:
Using Rule 1, we have 1
1
3
3
Sum opposite to 30o angle = a = b 3 = or c 2
=
2a
2
× 15 ×
3
=5
3 cm
and b = a
or c = 2a Side opposite 45o
Side opposite 45o
Side opposite 90o
a
a
2a
1
1
2
Properties 45o − 45o − 90o Triangle
1. Side opposite 45o =
=
=
=
1 2 2 2 1 2
×
×
2 2
1 2
times the hypotenuse
times the hypotenuse (Rationalizing the denominator)
times the hypotenuse
2 × hypotenuse
Therefore Property 1: The length of the side opposite 45o angle equals one-half the length of the hypotenuse times the square root of 2.
2. Side opposite 45o (Hypotenuse) 2a =
c=
o
2 × side opposite 45 angle
Therefore Property 2: The length of the hypotenuse equals the length of a side times the square root of 2.
SOLVED EXAMPLES 1. The length of the hypotenuse of a 30o − 60o − 90o triangle is 18 cm. Find the length of its other sides. Sol:
We know that the length of the side opposite 30o
=
1 2
× 18 =
=
1 2
× length of hypotenuse
9 cm
Length of the side opposite 60o angle =
3 2
c=
3 2
× 18 =
9 3 cm
2. The length of a side of a 30o − 60o − 90o triangle which is opposite 30o angle is 5 cm. Find the length of its other two sides. Sol:
By property of the 30o − 60o − 90o triangle, we have Using Rule 2, we have Hypotenuse of the triangle =
1 3
× 2×
Verification
According to Pythagoras Theorem
3 × 15 = 10 3 cm
a2
2
+b =
c2
Now, left hand side = a 2 + b 2 and right hand side = c 2
=
=
(5 3 )
(10 3 )
2
=
2
2
+ (15) =
30 0
75 + 225 = 300 … (1)
… (2)
From (1) and (2), we find that
LHS=RHS Hence verified *******************************************************
