Subject: Optional Maths
A triangle is a geometrically closed figure which is bounded by three lines and angles. It is categorized into two types as sides and angles.
We can categorize the triangle into following 3 types on the basis of sides:
We can categorize the triangle into following 3 types on the basis of angle:
A right-angled triangle which is taken into consideration is known as the reference angle. For example, Assuming \(\angle\) EFG as the reference angle. where angle E is known as remaining angle. So,opposite to the right angle is known as hypotenuse and it is denoted by h. The side opposite to the reference angle, the side is called perpendicular which is denoted by p and remaining side are known as the base and it is denoted by b ,
Here,
EF = hypotenuse(h)
EG=Perpendicular (p)
FG= base(b)
In trigonometry, most of the reference angles are represent by verious Greek alphabets. Some of them are given below:
alpha | \(\alpha\) |
beta | \(\beta\) |
gamma | \(\gamma\) |
delta | \(\delta\) |
We normally use these symbols for representing reference angles.
For example,
Here, \(\triangle\) EFG is a right angled triangle, right angled at F.
Let, ∠GEF = θ be the reference angles .
Then, side EG = hypotenuse ( h ) [\(\therefore\)It is opposite to right angle ]
side EF = perpendicular ( p ) [\(\therefore\)It is opposite to reference angle ]
side FG = base ( b ) [\(\therefore\)It is side opposite to remaining angle ]
Pythagoras Theoramem
Perpendicular, Base and Hypotenuse is established by a famous mathematician named Pythagoras. Pythagoras Theorem is a relation developed on the basis of the elements of right angled triangle.
Here, Pythagoras theorem states that the sum of squares of the perpendicular and base of a right-angled triangle is equal to the squares of the hypotenuse . In simple language , the square made on the hypotenuse of a right-angled triangle is always equal to the sum of squares on the remaining two sides.
Mathematically, (perpendicular}2 + (base)2 = (hypotenuse)2 [i.e. p2 + b2 = h2]
For example:
If perpendicular (p) = 4cm, base (b) = 3 cm, then hypotenuse (h) = ?
Solution:
Perpendicular (p) = 2 cm
Base (b) = 3 cm
Hypotenuse (h) = ?
By formulae
h2= p2 + b2
or, h2 =42 + 32
or, h2 = 16 + 9
or, h2 = \(\sqrt{25}\)
\(\therefore\) h = 5 cm.
Note: h2= p2 + b2 p2 = h2 - b2 , p = \(\sqrt{h^2 - b^2}\) b2= h2- p2, b = \(\sqrt{h^2 - p^2}\) |
Converse of Pythagoras Theorem
As we know that in a right-angled triangle , h2= p2 + b2 which is known as Pythagoras theorem .On the contrary, the converse of Pythagoras relation holds true i.e, if h2= p2 - b2 holds true then the triangle is the right angled . If the relation is false , the triangle is not theright-angled triangle . For example:
Given that hypotenuse (h) = 4 cm, base (b) = 3 cm and perpendicular (p) = 2 cm. Whether it is true or false?
Solution:
Hypotenuse (h) = 4 cm
Base (b) = 3 cm
Perpendicular (p) = 2 cm
By formulae,
h2 = p2 + b2
or, (4cm)2 = (2cm)2 + (3cm)2
or, 16cm2 = 4cm2 + 9cm2
or 16cm2 = 13cm2, which is false.
Define triangle?
A triangle is a geometrically closed figure which is bounded by three lines and angles .It is divided into two types they are sides and angles.
What is a reference angle ?
A right-angled triangle which is taken into consideration is known as the reference angle
What is hypotenuse ?
The side opposite to right angle is known as hypotenuse .
What is perpendicular ?
The side opposite to reference angle is known as perpendicular.
What is a base ?
The side except hypotenuse and the perpendicular is called base .
Find the unknown side from the given right angled triangles .
Solution;
Let, PQR be a right angled triangle where right angled at Q .
So, PQ = 3 cm
QR = 4 cm
PR = ?
We know, in right angled PQR ,
h2 = p2 + b2
so,( PR )2= ( PQ )2 + ( QR )2
(PR)2 = (3cm)2 + (4cm)2
(PR)2 = 9 cm + 12 cm
(PR)2 = 25 cm2
or, (PR )2= ( 5cm )2
∴ PR = 5 cm .
Hence , the unknown side PR = 5 cm .
ΔABC is a right angled triangle , right angled at c .
Solution ;
Here, AB = 13 cm
AC = 5 cm
BC = ?
We know, in right angled ΔABC ,
h2 = p2 + b2
or, ( AB )2= ( AC )2 + ( BC )2
or, ( 13 cm )2= (5 cm)2 + ( BC )2
or, 169 cm2 = 25 cm2 +( BC )2
or, 169 cm2 = 25 cm2 + ( BC )2
or, 169 cm2 - 25 cm2 = ( BC )2
or, 144 cm 2= ( BC )2
or, ( BC )2 = 144 cm 2
or, ( BC )2 = ( 12 cm )2
∴ BC = 12 cm
Hence, the unknown side BC = 12 cm.
Here, MNO is a right angled triangle, right angled at N. Find MO = ?
Solution :
So, MN = 1 cm
NO = 1 cm
MO = ?
We know ,
or, h2 = p2 + b2 (Since, MNO is a right angled triangle )
or, ( MO )2= ( MN )2 + ( NO )2
or, ( MO )2= (1 cm )2+ ( 1 cm )2
or, ( MO )2= 1 cm + 1 cm
or, ( MO )2= 2cm
∴ MO = \(\sqrt{2}\) cm
Hence, the reqired side, MO = \(\sqrt{2}\) cm .
Find p , b and h from the figures given below with the reference angles given .
Solution ;
Here, ΔABC is a right angled triangle right angled at A.
i.e. ∠BAC = 90 .
∠ACB = α ( reference angle )
∴ Side AB = perpendicular ( p )
Side BC = Hypotenuse ( h )
Side AC =base ( b )
Determine whether the triangle is right angled triangle or not ?
1)
2)
1 .
Solution ;
Here, PT = 3 cm
PO = 2cm
OT = 2 cm
Now, for ΔPOT to be a right angled triangle ,
h2 = p 2+ b2
or, ( PT )2= ( PO )2 + ( TO )2 ( PT is the longest side )
or, ( 3 cm )2 = ( 2 cm )2+ ( 2 cm )2
or, 9 cm2 = 4 cm2 + 4 cm2
or, 9 cm2 = 8 cm2 ( which is false )
Hence, Δ POT is not a right angled triangle .
2.
Solution :
Here, AU = 5 cm
US = 3 cm
SA = 4 cm
Now, for the ΔUSA to be a right angled triangle ,
h2 = p2 + b2
As, AU = 5 cm, lets consider it to be the hypotenuse.
[\(\therefore\) The longest side is the hypotenuse]
Now, (UA)2 = (US)2 + (SA)2
(5 cm)2 = (3 cm)2 + (4 cm)2
25 cm2 = 9cm2 +16 cm2
25 cm2 = 25 cm2 which is true.
Hence, \(\triangle\)USA is a right angled triangle.
The figure is Δ ABC right angled triangle. Find sinθ and cos∝.
Here p= 3cm, b = 4 cm
Solution
Here, in the right-angled Δ ABC, ACD = 90
So, h2 = P2 + b2
(AB)2 = (AC)2 + (BC)2
(AB)2 = (3cm)2 + (4 cm)2
(AB)2 = 9 cm2 + 16 cm2
(AB)2 = 25 cm2
(AB0 = 5 cm
Now, taking θ as the reference angle,
sinθ = \(\frac{p}{h}\)
= \(\frac{BC}{AB}\) = \(\frac{3}{4}\)
Again, takiing ∝ as the reference angle,
cos∝ = \(\frac{b}[h}\) = \(\frac{BC}{AB}\) = \(\frac{3}{4}\)
What is a Pythagoras Theorem?
Pythagoras Theorem is a relation developed on the basis of the elements of right angled triangle. It is established by a famous mathematician named Pythagoras. The elements of a right-angled triangle are perpendicular, base and hypotenuse.
Find p , b and h from the figures given below with the reference angles given.
Solution ;
Here, ΔQRS is a right angled triangle right angled at R.
i.e. ∠QRS = 90 .
∠QSR = α ( reference angle )
∴ Side QR = perpendicular ( p )
Side QS = Hypotenuse ( h )
Side SR =base ( b )
Again,
we know that,
h2 = p2 + b2
h2 = (3 cm)2 + (4 cm)2
h2 = 9 cm + 12 cm
h2 = 25 cm
h2 = (5 cm)2
h = 5
The figure is Δ ABC right angled triangle. Find sinθ and cos∝.
Here p= 5cm, b = 12 cm
Solution
Here, in the right-angled Δ ABC = 90
So, h2 = P2 + b2
(AB)2 = (AC)2 + (BC)2
(AB)2 = (5cm)2 + (12 cm)2
(AB)2 = 25 cm2 + 144 cm2
(AB)2 = 169 cm2
AB = 13 cm
Now, taking θ as the reference angle,
sinθ = \(\frac{p}{h}\)
= \(\frac{BC}{AB}\) = \(\frac{3}{4}\)
Again, takiing ∝ as the reference angle,
cos∝ = \(\frac{b}{h}\)
= \(\frac{BC}{AB}\) = \(\frac{3}{4}\)
The figure is Δ ABC right angled triangle. Find sinθ and cos∝.
Here p= 5cm, b = 12 cm
Solution
Here, in the right-angled Δ ABC = 90
So, h2 = P2 + b2
(AB)2 = (AC)2 + (BC)2
(AB)2 = (25cm)2 + (12 cm)2
(AB)2 = 25 cm2 + 144 cm2
(AB)2 = 169 cm2
(AB)2 = (13 cm)2
AB = 13 cm.
The figure is Δ ABC right angled triangle.
Here p= 5cm, b = 12 cm
Solution
Here, in the right-angled Δ ABC = 90
So, h2 = P2 + b2
(AB)2 = (AC)2 + (BC)2
(AB)2 = (25cm)2 + (12 cm)2
(AB)2 = 25 cm2 + 144 cm2
(AB)2 = 169 cm2
(AB)2 = (13 cm)2
AB = 13 cm.
Here, ABC is a right angled triangle , right angled at A.
AB = 2 cm
BC = 2cm
AC = ?
Solution
We know ,
h2 = p2 + b2( Since ΔABC is a right angled triangle)
(AB)2 = (AB)2 + (BC)2
(AB)2 = (2cm)2 + (2cm)2
(AB)2 = 4cm2
AB = 4 cm ans.
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