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Maths-

General

Easy

Question

Consider a triangle ABC. BE and CF are the medians drawn through the angular points B and C respectively and G is the centroid of the $capital delta$ ABC.
Statement – 1: If the points, A, E, G, F are concyclic then $angle A$ must be acute. because
Statement – 2: The medians of a triangle divide triangle into area wise two equal parts.

Statement – 1 is True, Statement – 2 is a correct explanation for Statement – 1
Statement – 1 is True, Statement – 2 is True, Statement – 2 is not a correct explanation for Statement – 1
Statement – 1 is True, Statement – 2 is False
Statement – 1 is False, Statement – 2 is True

The correct answer is: Statement – 1 is True, Statement – 2 is True, Statement – 2 is not a correct explanation for Statement – 1

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Which one of these is a derived physical quantity?

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Which of the following numbers is a perfect number?

Which of the following numbers is a perfect number?

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Which of the following is even?

Which of the following is even?

$c o s invisible function application e c to the power of 2 end exponent 60 to the power of ring operator end exponent s e c to the power of 2 end exponent invisible function application 30 to the power of ring operator end exponent c o s invisible function application 0 to the power of ring operator end exponent s i n invisible function application 45 to the power of ring operator end exponent c o t to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent t a n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent equals$

$c o s invisible function application e c to the power of 2 end exponent 60 to the power of ring operator end exponent s e c to the power of 2 end exponent invisible function application 30 to the power of ring operator end exponent c o s invisible function application 0 to the power of ring operator end exponent s i n invisible function application 45 to the power of ring operator end exponent c o t to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent t a n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent equals$

$4 open parentheses s i n to the power of 4 end exponent invisible function application 30 to the power of ring operator end exponent plus c o s to the power of 4 end exponent invisible function application 60 to the power of ring operator end exponent close parentheses plus fraction numerator 2 over denominator 3 end fraction open parentheses s i n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent minus c o s to the power of 2 end exponent invisible function application 45 to the power of ring operator end exponent close parentheses plus fraction numerator 1 over denominator 2 end fraction t a n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent equals$

$4 open parentheses s i n to the power of 4 end exponent invisible function application 30 to the power of ring operator end exponent plus c o s to the power of 4 end exponent invisible function application 60 to the power of ring operator end exponent close parentheses plus fraction numerator 2 over denominator 3 end fraction open parentheses s i n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent minus c o s to the power of 2 end exponent invisible function application 45 to the power of ring operator end exponent close parentheses plus fraction numerator 1 over denominator 2 end fraction t a n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent equals$

parallel

cot $theta$ + tan $theta$ =

Hence option 2 is correct

cot $theta$ + tan $theta$ =

Hence option 2 is correct

$open parentheses 1 plus c o t to the power of 2 end exponent invisible function application theta close parentheses left parenthesis 1 minus c o s invisible function application theta right parenthesis left parenthesis 1 plus c o s invisible function application theta right parenthesis equals$

Hence option 2 is correct

$open parentheses 1 plus c o t to the power of 2 end exponent invisible function application theta close parentheses left parenthesis 1 minus c o s invisible function application theta right parenthesis left parenthesis 1 plus c o s invisible function application theta right parenthesis equals$

Hence option 2 is correct

If $s i n invisible function application A equals fraction numerator 1 over denominator 3 end fraction$ , then cos A cosecA + tan A sec A =

If $s i n invisible function application A equals fraction numerator 1 over denominator 3 end fraction$ , then cos A cosecA + tan A sec A =

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