#### Question

In rectangle ABCD AB = 20cm ∠BAC = 60° BC, calculate side BC and diagonals AC and BD.

#### Solution

We have drawn the following figure

Since *ABCD* is a rectangle

Therefore

`∠ABC = ∠BCD = 90^@`

Now, consider ΔABC

We know that sum of all the angles of any triangle is 180º

Therefore,

∠BAC + ∠ABC + ∠ACB = 180° .....(1)

Now by substituting the values of known angles ∠BAC and ∠ABC inn equation (1)

We get

`60^@ + 90^@ + ∠ACB = 180^@`

`=> 150^@ + ∠ACb = 180^@`

`=> ∠ACB = 1806@ - 150^@`

`=> ∠ACb = 30^@`

Now in ΔABC

We know that

`cos A = cos 60^@`

`=> (AB)/(AC) = 60^@` ....(2)

Now we have,

`AB = 20 cm and cos 60^@ = 1/2`

Therefore by substituting above values in equation (2)

We get,

`cos A = cos 60^@`

`=> 20/(AC) = 1/2`

Now by cross multiplying we get,

`20 xx 2 = 1 xx AC`

=> 40 = AC

`=> AC = 40` ....(3)

Now in ΔABC

We know that,

`sin A = sin 60^@`

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

Now we have from equation (3),

`AC = 40 cm and sin 60^@ = sqrt3/2`

Therefore by substituting above values in equation (4)

We get,

`sin A = sin 60^@`

`=> (BC)/40 = sqrt3/2`

Now by cross multiplying we get,

`BC xx 2 = sqrt3 xx 40`

`=> BC = (sqrt3 xx 40)/2`

`=> BC = 20sqrt3`

Therefore

`BC = 20sqrt3 m` .....(5)

Since *ABCD* is a rectangle

Therefore,

AB = CD = 20 cm ....(6)

And

`BC = AD = 20sqrt3` cm .....(7)

Now in ΔBCD

We know that,

`tan B = (CD)/(BC)`

Now by substituting the values of sides from equation (6) and (7)

We get,

`tan B = 20/(20sqrt3)`

`=> tan B = 1/sqrt3`

Since

`tan 30^@ = 1/sqrt3`

Therefore ∠B = 30°

That is in ΔBCD

∠DBC = 30° ....(8)

Now in ΔBCD

We know that,

`cos B = (BC)/(BD)`

From equation (7)and (8)

`=> cos 30^@ = (20sqrt3)/(BD)`

Since

`cos 30^@ = sqrt3/2`

Therefore,

`sqrt3/2 = (20sqrt3)/(BD)`

Now by cross multiplying we get,

`sqrt3 xx (BD) = 20sqrt3 xx 2`

`=> sqrt3 xx BD = 40sqrt3`

`=> BD = (40sqrt3)/sqrt3`

=> BD = 4

Therfore

BC= 40 cm ...(9)

Hence from equation (3), (5) and (9)

`AC = 40 cm, BC = 20sqrt3 cm, BD = 40 cm`