Q. 14.1( 50 Votes )

# In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

Answer :

__For first column:__

cosθ = 35/37

Adjacent side= 35,

Hypotenuse = 37

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Opposite side^{2} = Hypotenuse^{2} - Adjacent^{2}

= 37^{2} - 35^{2}

= 1369 – 1225

Opposite side^{2} = 144

Opposite side = 12

__For second column:__

Opposite side = 11

Hypotenuse = 61

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Adjacent^{2} = Hypotenuse^{2} - Opposite side^{2}

= 61^{2} - 11^{2}

= 3721 – 121

Adjacent^{2} = 3600

Adjacent side= 60

__For third column:__

Opposite side = 1

Adjacent side= 1

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

= 1 + 1

Hypotenuse^{2} = 2

Hypotenuse = √2

__For fourth column:__

Opposite side = 1

Hypotenuse = 2

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Adjacent^{2} = Hypotenuse^{2} - Opposite side^{2}

= 2^{2} - 1^{2}

= 4 – 1

Adjacent^{2} = 3

Adjacent side= √3

__For fifth column:__

Adjacent side= 1

Hypotenuse = √3

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Opposite side^{2} = Hypotenuse^{2} - Adjacent^{2}

= (√3)^{2} - 1^{2}

= 3 – 1

Opposite side^{2} = 2

Opposite side = √2

__For sixth column:__

Opposite side = 21

Adjacent side= 20

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

= 21^{2} + 20^{2}

Hypotenuse^{2} = 841

Hypotenuse = 29

__For seventh column:__

Opposite side = 8

Adjacent side= 15

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

= 8^{2} + 15^{2}

Hypotenuse^{2} = 289

Hypotenuse = 17

__For eighth column:__

Opposite side = 3

Hypotenuse = 5

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Adjacent^{2} = Hypotenuse^{2} - Opposite side^{2}

= 5^{2} - 3^{2}

= 25 – 9

Adjacent^{2} = 16

Adjacent side= 4

__For ninth column:__

Opposite side = 1

Adjacent side= 2√2

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

= 1^{2} + (2√2)^{2}

Hypotenuse^{2} = 9

Hypotenuse = 3

Rate this question :

Find the values of –

2sin 30^{0} + cos 0^{0} + 3sin 90^{0}

Find the values of –

cos^{2}45^{0} + sin^{2}30^{0}

Find the values of –

cos 60^{0}× cos 30^{0} + sin60^{0} × sin30^{0}

In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

MHB - Math Part-II

Find the values of –

MHB - Math Part-II

Find the values of –

5sin 30^{0} + 3tan45^{0}

Fill in the blanks.

i.

ii.

iii.

MHB - Math Part-II