A right triangle or right-angled triangle is a triangle which contains a right angle, whereas the right angle is an angle measuring $90^{\circ}$. The following figure shows a right-angled triangle: |

A triangle is known as right triangle, if it has one angle as right angle.

The right angle in a right triangle is shown either by notation $90^{\circ}$ or by the symbol as shown in the figure above (mark in red color).

In a right triangle, $\angle C=90^{\circ}$. Let us assume that side opposite to $\angle C$ is "c", side opposite to $\angle A$ is "a" and side opposite to $\angle B$ is denoted by "b".

A theorem known as

**Pythagoras Theorem**is defined in right-angled triangle. It states that, "In a right triangle, square of the longest side or the hypotenuse is equal to the sum of squares of other two sides."

Pythagoras Theorem is expressed as:

Perimeter of right triangle is given by:

Trigonometric ratios are defined in right triangle as shown below:

Let us consider a right triangle $\bigtriangleup ACB$ right angled at C. If $\angle B$ is named as $\theta $.

Then, following rules hold:

$\sin \theta$ = $\frac{AC}{AB}$

$\cos \theta$ = $\frac{BC}{AB}$

$\tan \theta$ = $\frac{AC}{BC}$

$\csc \theta$ = $\frac{AB}{AC}$

$\sec \theta$ = $\frac{AB}{BC}$

$\cot \theta$ = $\frac{BC}{AC}$

Sine Rule:

Sine Rule:

**Cosine Rule:**

$90^{\circ}+90^{\circ}+90^{\circ}=270^{\circ}$

which is greater than $180^{\circ}$. Hence, an equilateral right triangle is not at all possible.

An acute triangle is one which has all its angles measuring less than $90^{\circ}$. But, in a right triangle, one angle must be $90^{\circ}$. So, an acute right triangle will not be practically possible, since a right triangle must be having one right angle and two acute angles.

An obtuse triangle is one which has at least one of its angles measuring greater than $90^{\circ}$. But, in a right triangle, one angle must be $90^{\circ}$. So, an obtuse triangle will also not be practically possible, since a right triangle must be having one right angle and two acute angles.

If we try to hypothetically construct a right triangle with an obtuse angle, then sum of all three angles will be greater than $180^{\circ}$. Hence, a right obtuse triangle is not possible.

Right triangles are everywhere in real life. They are all around us. There are many such examples. Few are as follows:

- If we place a ladder against the wall, it makes a right triangle with the wall.

- While finding distance or angle of elevation from a ship to a tall tower, it forms right angle.

- A sandwich is in the shape of right triangle.

- A triangle made by staircase and wall is always a right triangle.

There are many many more examples of right triangles which we come across in real life.

### Solved Examples

**Question 1:**The angle of elevation of a 10-meter high building from a fixed point is noted to be $60^{\circ}$. Find the distance of the point from the base of the building.

**Solution:**

**Step 1:**The following figure is obtained according to the question:

Let us assume that the distance of given point from the base of the building is x meter.

According to a property of right triangle, we have

$\tan \theta$ = $\frac{AB}{BC}$

$\tan 60^{\circ}$ = $\frac{10}{x}$

$\sqrt{3}$ = $\frac{10}{x}$

$x$ = $\frac{10}{\sqrt{3}}$

$x$ = $\frac{10}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

= $\frac{17.32}{3}$

= 5.8 meter (approx)

**Step 2:**The following figure is obtained according to the question:

Let us assume that the distance of given point from the base of the building is x meter.

According to a property of right triangle, we have

$\tan \theta$ = $\frac{AB}{BC}$

$\tan 60^{\circ}$ = $\frac{10}{x}$

$\sqrt{3}$ = $\frac{10}{x}$

$x$ = $\frac{10}{\sqrt{3}}$

$x$ = $\frac{10}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

= $\frac{17.32}{3}$

= 5.8 meter (approx)

**Question 2:**A ladder is leaning against the wall. Top of the ladder touches the wall at a point which is 4 meter high from the ground, while its bottom is 3 meter away from the base of wall. What is the length of the ladder?

**Solution:**

Here, AC is the ladder and $\bigtriangleup ABC$ is the right triangle.

By Pythagorean theorem, In $\bigtriangleup ABC$:

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

$(AC)^{2}=(4)^{2}+(3)^{2}$

$(AC)^{2}=16+9$

$(AC)^{2}=25$

$AC=\sqrt{25}$

$AC=5\ meter$

### Solved Examples

**Question 1:**Determine whether the following sides make a right triangle:

(a) 8, 6, 10

(b) 6, 7, 8

**Solution:**

**Square of the longest side = $10^{2}=100$**

(a)

(a)

Sum of squares of other two sides = $6^{2}+8^{2}$

= $36+64$ = 100

Square of the longest side = Sum of squares of other two sides.

Hence, these sides represent a right triangle.

**Square of the longest side = $8^{2}=64$**

(b)

(b)

Sum of squares of other two sides = $6^{2}+7^{2}$

= $36+49$ = 85

Square of the longest side $\neq $ Sum of squares of other two sides.

Hence, these sides do not represent a right triangle.

**Question 2:**Calculate the area and perimeter of a triangle whose length of base and height are 5 cm and 12 cm respectively.

**Solution:**

=

**$\frac{1}{2}$ x 5 x 12**

= 30

Area of right triangle is 30$cm^{2}$

$(Hypotenuse)^{2}=5^{2}+12^{2}$

$(Hypotenuse)^{2}=25+144$

= 169

$Hypotenuse=\sqrt{169}$

$Hypotenuse=13 cm$

Perimeter of the right triangle = 5 + 12 + 13

= 30

Perimeter of the right triangle is 30 cm