Solution of Right Angled Triangle

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

Right angled triangle ABC.

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 we have,

          b2 = a2 + c2 ……………… (i)

          A + C = 90° ……………… (ii)

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 right angled triangle.


 

Methods of solution of a right angled triangle:

(a) When two sides are given:

 (i)          Suppose 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)        Suppose 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.

(b) When one side and one acute angle are given:

 (i)          Suppose A and b are given.

Then, C = 90° - A

sinA = a/b

  a = b sinA

cosA = c/b

  c = b cosA

 (ii)        Suppose 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: 

In right angled triangle ABC, ∠B = 90°, ∠C = 45° and BC = √2

In DABC, B = 90°, C = 45° and BC = √2

Then, 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

(ii)        Figure: 

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

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

AB2 + BC2 = (√3)2 + (1)2

              = 3 + 1

              = 4

              = (2)2

              = AC2

  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°

 

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

In right angled triangle ABC, ∠C = 90°, ∠B = 30° and a = BC = 4√3 cms.

Solution:            

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

          We know, 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 = 8 cms, b = AC = 4 cms and A = 60°.


 


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

In triangle ABC, ∠B = 90°, a = BC = 2√3 and b = AC = 4.

Solution:

          In triangle ABC, B = 90°, a = BC = 2√3 and b = AC = 4.

          Now, cosC = BC/AC

          or,      cosC = 2√3/4 = √3/2

          or,      cosC = 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.

 

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

In triangle ABC, ∠A = 90°, b = 16 and c = 16.

Solution:

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

            a2 = b2 + c2

                  = (16)2 + (16)2

                  = 2 × (16)2

             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.

 

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

In triangle ABC, a = BC = 5√3, b = AC = 5 and c = AB = 10.

Solution:

          In triangle ABC, a = BC = 5√3, b = AC = 5 and c = AB = 10.

          AC2 + BC2 = (5)2 + (5√3)2

                           = 25 + 75

                          = 100 = (10)2 = AB2

            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°.

 

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

In triangle ABC, ∠B = 60°, ∠A = 30° and a = BC = 20.

Solution:

          In triangle ABC, B = 60°, A = 30° and a = BC = 20

          C = 180° - (A + B)

      = 180° - (30° + 60°)

      = 180° - 90°

      = 90°

Now, 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.


 

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

In triangle ABC, ∠B = 90°, c = AB = 4√3 cms and a = BC = 12 cms.

Solution:

          In triangle ABC, B = 90°, c = AB = 4√3 cms and a = BC = 12 cms.

          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 cms.

            A = 60°, C = 30° and b = AC = 8√3 cms.

 

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

In triangle ABC, a = BC = 3, b = AC = 4 and c = AB = 5.

Solution:

          In triangle ABC, a = BC = 3, b = AC = 4 and c = AB = 5

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

            a2 + b2 = c2

            C = 90°

          Now, cosB = a/c = 3/5 = cos54°

            B = 54°

          And, A = 90° - B

            = 90° - 54°

            = 36°

            A = 36°, B = 54° and C = 90°

 

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