In right angled triangle ABC. ∠C = 90°, ∠B = 60°. AB = 15 units. Find remaining angles and sides.

#### Solution 1

We are given the following triangle with related information

It is required to find ∠A, ∠C and length of sides AC and BC

ΔABC is right angled at C

Therefore,

`∠C = 90^@`

Now we know that sum of all the angles of any triangle is `180^@`

Therefore

`∠A + ∠B + ∠C = 180^@`

Now by substituting the values of known angles and in equation (1)

We get,

`∠A + 60^@ + 90^2 = 180^@`

Therefore

`∠A + 150^@ = 180^@`

`=> ∠A = 180^@ - 150^@`

`=> ∠A = 30^@`

Therefore,

`∠A = 30^@`

Now

We know that,

`cos B = cos 60^@`

`=> (BC)/(AB) = cos 60^@`

Now we have,

*AB *=15 units and `cos 60^@ = 1/2`

Therefore by substituting above values in equation (2)

We get,

`cos B = cos 60^@`

`=> (BC)/(AB) = cos 60^@`

`=> (BC)/15 = 1/2`

Now by cross multiplying we get,

`(BC)/15 = 1/2`

`=> 2 xx BC = 15 xx 1`

`=> BC = 15/2`

=> BC = 7.5

Therefore

BC = 7.5 units ....(3)

Now

We know that

`sin B = sin 60^@`

`=> (AC)/(AB) = sin 60^@` .........(4)

Now

We know that,

AB=15 units and

#### Solution 2

We are given the following triangle with related information

It is required to find ∠A, ∠C and length of sides AC and BC

ΔABC is right angled at C

Therefore,

`∠C = 90^@`

Now we know that sum of all the angles of any triangle is `180^@`

Therefore

`∠A + ∠B + ∠C = 180^@`

Now by substituting the values of known angles and in equation (1)

We get,

`∠A + 60^@ + 90^2 = 180^@`

Therefore

`∠A + 150^@ = 180^@`

`=> ∠A = 180^@ - 150^@`

`=> ∠A = 30^@`

Therefore,

`∠A = 30^@`

Now

We know that,

`cos B = cos 60^@`

`=> (BC)/(AB) = cos 60^@`

Now we have,

*AB *=15 units and `cos 60^@ = 1/2`

Therefore by substituting above values in equation (2)

We get,

`cos B = cos 60^@`

`=> (BC)/(AB) = cos 60^@`

`=> (BC)/15 = 1/2`

Now by cross multiplying we get,

`(BC)/15 = 1/2`

`=> 2 xx BC = 15 xx 1`

`=> BC = 15/2`

=> BC = 7.5

Therefore

BC = 7.5 units ....(3)

Now

We know that

`sin B = sin 60^@`

`=> (AC)/(AB) = sin 60^@` .........(4)

Now

We know that,

`AB=15 units and sin 60^@ = sqrt3/2`

Therefore by substituting above values in equation (4)

We get,

`sin B = sin 60^@`

`=> (AC)/(AB) = sin 60^@`

`=> (AC)/15 = sqrt3/2`

Now by cross multiplying we get,

`=> 2 xx AC = sqrt3 xx 15`

`=> AC = (sqrt3 x 15)/2`

`=> AC = 15/2 sqrt3`

Therefore,

AC = 15/2 sqrt3` units`

Hence

`A = 30^@`

`BC = 7.5untis`

`AC = 15/2 sqrt3` units