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:

**Equilateral triangle**

A triangle having all sides equal is known as an equilateral triangle. In figure (i) we have observed a triangle with all 3 sides are equal.**Isosceles triangle**

A triangle having any two sides equal is known as an isosceles triangle. In figure (ii), we can see a triangle with two sides equal.**Scalene triangle**

A triangle having none of the sides equal is known as an isosceles triangle. In figure (iii), we can see a triangle with no sides equal to each other.

We can categorize the triangle into following 3 types on the basis of angle:

**Acute angled triangle**A triangle having an angle less than 90 degrees is known as acute angled triangle. In figure (i), we have seen a triangle having less than 90.**Obtuse angle triangle**

A triangle having an angle more than 90 degrees is known as obtuse angled triangle. In figure (ii), one of the angles is more than 90 degree.**Right angle triangle**A triangle having an angle exactly 90 degree is known as right angled triangle. In figure (iii), one of angle is 90.

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. p^{2} + b^{2} = h^{2}]

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

h^{2}= p^{2} + b^{2}

or, h^{2} =4^{2} + 3^{2}

or, h^{2} = 16 + 9

or, h^{2} = \(\sqrt{25}\)

\(\therefore\) h = 5 cm.

Note: h p b |

**Converse of Pythagoras Theorem**

As we know that in a right-angled triangle , h^{2}= p^{2} + b^{2} which is known as Pythagoras theorem .On the contrary, the converse of Pythagoras relation holds true i.e, if h^{2}= p^{2} - b^{2} 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,

h^{2} = p^{2} + b^{2}

or, (4cm)^{2} = (2cm)^{2} + (3cm)^{2}

or, 16cm^{2} = 4cm^{2} + 9cm^{2}

or 16cm^{2} = 13cm^{2}, which is false.

- A geometrical closed figure bounded by three lines and angles is called triangle.
- We represent the reference angles by various Greek alphabets .
- Here we use p , b and h to represent perpendicular , base , hypotenuse respectively .They represent the sides so the alphabets p, b and h always should be in lower case.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

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 ,

h^{2 }= p^{2 }+ b^{2}

so,( PR )^{2}= ( PQ )^{2} + ( QR )^{2}

(PR)^{2} = (3cm)^{2} + (4cm)^{2 }

(PR)^{2} = 9 cm + 12 cm

(PR)^{2} = 25 cm^{2}

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 ,

h^{2} = p^{2} + b^{2}

or, ( AB )^{2}= ( AC )^{2} + ( BC )^{2}

or, ( 13 cm )^{2}= (5 cm)^{2} + ( BC )^{2}

or, 169 cm^{2 }= 25 cm^{2} +( BC )^{2}

or, 169 cm^{2 }= 25 cm^{2} + ( BC )^{2}

or, 169 cm^{2 }- 25 cm^{2} = ( 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, h^{2 }= p^{2} + b^{2} (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 ,

h^{2} = p ^{2}+ b^{2}

or, ( PT )^{2}= ( PO )^{2} + ( TO )^{2} ( PT is the longest side )

or, ( 3 cm )^{2 }= ( 2 cm )^{2}+ ( 2 cm )^{2}

or, 9 cm^{2} = 4 cm^{2} + 4 cm^{2}

or, 9 cm^{2 }= 8 cm^{2} ( 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 ,

h^{2} = p^{2} + b^{2}

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 cm^{2} = 9cm^{2} +16 cm^{2}

25 cm^{2} = 25 cm^{2} 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, h^{2} = P^{2} + b^{2}

(AB)^{2} = (AC)^{2} + (BC)^{2}

(AB)^{2} = (3cm)^{2} + (4 cm)^{2}

(AB)^{2} = 9 cm^{2} + 16 cm^{2}

^{ }(AB)^{2} = 25 cm^{2}

^{ }(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,

h^{2} = p^{2} + b^{2 }

h^{2} = (3 cm)^{2 }+ (4 cm)^{2}

h^{2} = 9 cm + 12 cm

h^{2} = 25 cm

h^{2} = (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, h^{2} = P^{2} + b^{2}

(AB)^{2} = (AC)^{2} + (BC)^{2}

(AB)^{2} = (5cm)^{2} + (12 cm)^{2}

(AB)^{2} = 25 cm^{2} + 144 cm^{2}

^{ }(AB)^{2} = 169 cm^{2}

^{ }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, h^{2} = P^{2} + b^{2}

(AB)^{2} = (AC)^{2} + (BC)^{2}

(AB)^{2} = (25cm)^{2} + (12 cm)^{2}

(AB)^{2} = 25 cm^{2} + 144 cm^{2}

^{ }(AB)^{2} = 169 cm^{2}

^{ }(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, h^{2} = P^{2} + b^{2}

(AB)^{2} = (AC)^{2} + (BC)^{2}

(AB)^{2} = (25cm)^{2} + (12 cm)^{2}

(AB)^{2} = 25 cm^{2} + 144 cm^{2}

^{ }(AB)^{2} = 169 cm^{2}

^{ }(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 ,

h^{2} = p^{2 } + b^{2}( Since ΔABC is a right angled triangle)

(AB)^{2} = (AB)^{2} + (BC)^{2}

(AB)^{2} = (2cm)^{2} + (2cm)^{2}

(AB)^{2} = 4cm^{2}

AB = 4 cm ans.

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