Solution of Right Angled Triangle

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 h² = p² + b², we have

b² = a² + c²

 

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, b² = a² + c².

∴  b = √(a² + c²)

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

At last, C = 90° - A.

(ii)     When, a and b are given.

Then, c² = b² – a².

∴  c = √(b² – a²)

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,

AB² + BC² = (√3)² + (1)²

                   = 3 + 1

                   = 4

                   = (2)²

                   = AC²

Since, AB² + BC² = AC²

∴  ∠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.

 

∴   a² = b² + c²

          = (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,

AC² + BC² = (5)² + (5√3)²

                   = 25 + 75

                   = 100 = (10)² = AB²

Since,  AC² + BC² = AB²

∴  ∠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,

a² + b² = 3² + 4² = 9 + 16 = 25 = 5² = c²

i.e.  a² + b² = c²

∴  ∠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.

 

 

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