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3 years ago

1-sin60÷cos60=1-Tan30÷1+Tan30

Mathematics

1 answer:

ipn [44]3 years ago

8 0

Step-by-step explanation:

LHS=(1-sin60)/cos60

=(1-√3÷2)/1÷2

=2(1-√3÷2)

=2-√3

RHS=(1-tan30)/(1+tan30)

={1-(1÷√3)}/{1+(1÷√3)}

={(√3-1)/√3}/{(√3+1)/√3}

=(√3-1)/(√3+1)

={(√3-1)(√3-1)}/{(√3+1)(√3-1)}

=(3-√3-√3+1)/(3-1)

=(4-2√3)/2

=2-√3

Therefore LHS=RHS

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krok68 [10]

Let x be the amount of money, you fund in Fund A and y be the amount of mone yyou fund in Fund B.

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2 years ago

a square field has a side 150 m long .There is a road 5 m wide all around the park outside it .Find the cost of constructing the

alisha [4.7K]

Answer:

Cost to construct the road = 775000 rupees

Step-by-step explanation:

Square field has a side = 150 m

Area of the field = (Side)²

= (150)²

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Similarly, width of the park with road around the field = 160 m

Area of the park with road = (Side)²

= (160)²

= 25600 m²

Area of road = Area of the park with road - Area of the park

= 25600 - 22500

= 3100 m²

Since, cost of constructing the road = 250 rupees per square meter

Cost to construct the road with the given area = 250 × 3100

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2 years ago

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nataly862011 [7]

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