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# In Rectangle Abcd Ab = 20cm ∠Bac = 60° Bc, Calculate Side Bc and Diagonals Ac and Bd. - CBSE Class 10 - Mathematics

ConceptTrigonometric Ratios of an Acute Angle of a Right-angled Triangle

#### 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

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#### APPEARS IN

Solution In Rectangle Abcd Ab = 20cm ∠Bac = 60° Bc, Calculate Side Bc and Diagonals Ac and Bd. Concept: Trigonometric Ratios of an Acute Angle of a Right-angled Triangle.
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