**Solution of Right
Angled Triangle**

A triangle has three sides and three angles. Angles and sides
together are said to be the six parts of a triangle.

********************

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In a right-angled triangle, one angle is 90°, so there are 5
parts unknown.

Under a given condition, the finding of the unknown parts of a right
angled triangle is known as the **solution
of a right-angled triangle**.

Let ABC be a right-angled triangle, right-angled at B. We denote its angles ∠CAB, ∠ABC, and ∠BCA by A, B, and C respectively.

Here, B = 90°.

The sides opposite to angles A, B, and C are denoted by a, b and
c respectively. So, BC = a, CA = b and AB = c.

Then,

By Pythagoras relation h^{2} = p^{2} + b^{2},
we have

b^{2} = a^{2} + c^{2}

Again, by the sum of two acute angles,

A + C = 90°

If we know one side and any one of the remaining four parts, we
can find the other parts of the right-angled triangle. This process of
calculating the unknown parts is known as the **solution of a right-angled triangle**.

**Methods of Solving a
Right Angled Triangle**

**- When two sides of a
right-angled triangle are given**

(i)
__When a and c, are
given__.

Then, b^{2} = a^{2} + c^{2}.

∴ b = √(a^{2 }+ c^{2})

Again, sinA = a/b and so A is found.

At last, C = 90° - A.

(ii)
__When, a and b are
given__.

Then, c^{2} = b^{2} – a^{2}.

∴ c = √(b^{2} – a^{2})

Again, sinA = a/b and it gives A.

At last, C = 90° - A.

**- When one side and one
acute angle are given**

(i)
__When A and b are
given__.

Then, C = 90° - A

sinA = a/b

∴ a = b sinA

cosA = c/b

∴ c = b cosA

(ii)
__When, A and a are
given__.

Then, C = 90° - A

sinA = a/b

∴ b = a/sinA = a cosecA

cosA = c/b

∴ c = b cosA

**Worked Out Examples**

**Example 1:** Find the unknown angles and the sides of the following right angled
triangles:

**Solution:**

(i)
Figure: Right angled triangle ABC

In DABC, ∠B = 90°, ∠C = 45° and BC = √2

∠A = 90° – ∠C = 90° – 45° = 45°

Now,

tan45° = AB/BC

or, 1
= AB/√2

or, AB
= √2

Again,

cos45° = BC/AC

or, 1/√2
= √2/AC

or, AC
= 2

∴ ∠A = 45°, AB = √2 and AC = 2
Ans.

(ii) Figure: Triangle ABC

In triangle ABC, AB = √3, BC = 1 and AC = 2

Now,

AB^{2} + BC^{2} =
(√3)^{2} + (1)^{2}

=
3 + 1

=
4

=
(2)^{2}

=
AC^{2}

Since, AB^{2} + BC^{2} = AC^{2}

∴ ∠B = 90°

Now,

sinC = AB/AC

or,
sinC = √3/2

or,
sinC = sin60°

∴ C = 60°

And, ∠A = 90° –
C = 90° – 60° = 30°

∴ ∠B = 90°, ∠C = 60°, and ∠A = 30° Ans.

**Example 2:** Solve the right angled triangle ABC in which ∠C = 90°, ∠B = 30° and a = 4√3 cms.

**Solution:** Here,

In DABC,
∠C = 90°, ∠B = 30° and a = BC = 4√3
cms.

∠A = 90° – C = 90° – 30° = 60°

Now,

cos30° = BC/AB

or, √3/2 = 4√3/AB

or, AB√3 = 8√3

or, AB = 8√3/√3 = 8 cms.

Again,

sin30° = AC/AB

or, 1/2 = AC/8

or, 2AC = 8

or, AC = 8/2 = 4 cms.

∴ c = AB = 8cms, b = AC =
4cms, and ∠A = 60° Ans.

**Example 3:** In a right angled triangle ABC, if ∠B = 90°, a = 2√3 and b =
4, find the remaining angles and the sides.

**Solution:** Here,

In triangle ABC,

∠B = 90°

a = BC = 2√3

b = AC = 4

Now,

cosC = BC/AC = 2√3/4 = √3/2 = cos30°

∴ C = 30°

And,

∠A = 90° – C = 90° – 30° = 60°

Again,

sin30° = AB/AC

or, 1/2 = AB/4

or, 2AB = 4

or, AB = 4/2 = 2

∴ ∠A = 60°, ∠C = 30°, and c = AB =
2 Ans.

**Example 4:** Solve the triangle ABC if ∠A = 90°, b = 16 and c = 16.

**Solution:** Here,

In triangle ABC,

∠A = 90°

b = AC = 16

c = AB = 16

Since ∠A = 90°, so a is the
hypotenuse.

∴ a^{2} =
b^{2} + c^{2}

= (16)^{2} + (16)^{2}

= 2 × (16)^{2}

or, a = 16√2

Now,

tanB = AC/AB = b/c = 16/16 = 1 = tan45°

∴ B = 45°

And,

∠C = 90° – B = 90° – 45° = 45°

∴ ∠B = 45°, ∠C = 45° and a = BC = 16√2
Ans.

**Example 5:** Solve the triangle ABC if a = 5√3, b = 5 and c = 10.

**Solution:** Here,

In triangle ABC,

a = BC = 5√3

b = AC = 5

c = AB = 10

Now,

AC^{2} + BC^{2} =
(5)^{2} + (5√3)^{2}

=
25 + 75

=
100 = (10)^{2} = AB^{2}

Since, AC^{2} + BC^{2}
= AB^{2}

∴ ∠C = 90°

Now,

cosA = AC/AB = 5/10 = ½ = cos60°

∴ A = 60°

And,

∠B = 90° – A = 90° – 60° = 30°

∴ ∠C = 90°, ∠B = 30°, and ∠A = 60° Ans.

**Example 6:** Solve a triangle ABC if ∠B = 60°, ∠A = 30° and a = 20.

**Solution:** Here,

In triangle ABC,

∠B = 60°

∠A = 30°

a = BC = 20

Now,

∠C = 180° – (∠A + ∠B)

= 180° – (30° + 60°)

= 180° – 90°

= 90°

And,

sin30° = BC/AB

or, 1/2 = 20/AB

or, AB = 40

Again,

cos30° = AC/AB

or, √3/2 = AC/40

or, 2AC = 40√3

or, AC = 40√3/2 = 20√3

∴ ∠C = 90°, c = AB = 40 and b = AC = 20√3 Ans.

**Example 7:** In a right angled triangle the sides containing the right angle are
4√3cm and 12cm. Find the hypotenuse and acute angles.

**Solution:** Here,

In triangle ABC,

∠B = 90°

c = AB = 4√3cm.

a = BC = 12cm.

Now,

tanC = AB/BC = 4√3/12 = 1/√3 = tan30°

∴ C = 30°

And,

∠A = 90° – C = 90° – 30° = 60°

Again,

sinC = AB/AC

or, sin30° = 4√3/AC

or, ½ = 4√3/AC

or, AC = 8√3 cm.

∴ ∠A = 60°, ∠C = 30°, and hypotenuse =
AC = 8√3cm. Ans.

**Example 8:** Solve the triangle ABC if a = 3, b = 4 and c = 5 [Take cos54° =
3/5]

**Solution:**Here,

In triangle ABC,

a = BC = 3

b = AC = 4

c = AB = 5

Now,

a^{2} + b^{2} = 3^{2} + 4^{2} =
9 + 16 = 25 = 5^{2} = c^{2}

i.e. a^{2} + b^{2}
= c^{2}

∴ ∠C = 90°

Again,

cosB = a/c = 3/5 = cos54°

∴ B = 54°

And, ∠A = 90° – B

= 90° – 54°

=
36°

∴ ∠A = 36°, ∠B = 54° and ∠C = 90° Ans.

If you have any questions or problems regarding the **Solution of Right Angled Triangle**, you can ask here, in the comment section below.

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