If $4 n alpha equals pi comma$ then the numerical value of $tan space alpha tan space 2 alpha tan space 3 alpha horizontal ellipsis horizontal ellipsis tan space left parenthesis 2 n minus 1 right parenthesis alpha$ is equal to

$sin space 20 to the power of ring operator open parentheses 4 plus sec space 20 to the power of ring operator close parentheses$ =

In the figure $P Q greater than P R comma$ QS and RS are the bisectors of $stack ∣ Q with _ below$ and $stack R with _ below$ respectively then

If AB=AC and $A C D equals 115 to the power of ring operator end exponent$ then $stack B with _ below A C equals$

Here we used the concept of linear pair. When two lines cross at one point, a linear pair of angles is created. If the angles follow the intersection of the two lines in a straight line, they are said to be linear. A linear pair's total angles are always equal to 180°. So here the angle A is 50 degree.

If AB=AC and $A C D equals 115 to the power of ring operator end exponent$ then $stack B with _ below A C equals$

In the figure $A equals 30 to the power of ring operator end exponent & E B equals 50 to the power of ring operator end exponent$ then ACD=

Here we used the concept of linear pair and exterior angle property. When two lines cross at one point, a linear pair of angles is created. If the angles follow the intersection of the two lines in a straight line, they are said to be linear. A linear pair's total angles are always equal to 180°. So here the angle BCD is 110 degrees.

In the figure $A equals 30 to the power of ring operator end exponent & E B equals 50 to the power of ring operator end exponent$ then ACD=

$open parentheses sin to the power of 8 space 75 to the power of ring operator minus cos to the power of 8 space 75 to the power of ring operator close parentheses$ =

If $tan space alpha equals fraction numerator 1 over denominator square root of x open parentheses x squared plus x plus 1 close parentheses end root end fraction comma tan space beta equals fraction numerator square root of x over denominator square root of x squared plus x plus 1 end root end fraction$ and $tan space gamma equals square root of x to the power of negative 3 end exponent plus x to the power of negative 2 end exponent plus x to the power of negative 1 end exponent end root$ then $alpha plus beta$ =

If $tan space left parenthesis pi cos space theta right parenthesis equals cot space left parenthesis pi sin space theta right parenthesis comma 0 less than theta less than fraction numerator 3 pi over denominator 4 end fraction$ then $sin space open parentheses theta plus pi over 4 close parentheses equals$

The minimum value $m space equals bold space 2 cos space theta plus fraction numerator 1 over denominator sin invisible function application theta end fraction plus square root of 2 tan space theta space text in end text open parentheses 0 comma pi over 2 close parentheses$ is

In the adjoining figure, $stack left floor a with _ below equals left floor b$ and $left floor c equals left floor d comma$ then $left floor b plus left floor c equals$

In the figure, $a equals 4 x to the power of ring operator end exponent$ and $b equals left parenthesis 2.5 x plus 24 right parenthesis to the power of ring operator end exponent comma$ then the value of x =

Here we used the concept of angles on same side of transversal. When two lines cross at one point, a linear pair of angles is created. If the angles follow the intersection of the two lines in a straight line, they are said to be linear. A linear pair's total angles are always equal to 180°. So here the value of x is 16.

In the figure, $a equals 4 x to the power of ring operator end exponent$ and $b equals left parenthesis 2.5 x plus 24 right parenthesis to the power of ring operator end exponent comma$ then the value of x =

In the following figure, the value of x =

Using information given in the following figure, the value of x and y is

Here we used the concept of linear pair and angle sum property of a triangle. When two lines cross at one point, a linear pair of angles is created. If the angles follow the intersection of the two lines in a straight line, they are said to be linear. A linear pair's total angles are always equal to 180°. So here the angles are 65°and 110°.

Using information given in the following figure, the value of x and y is

In figure, the sides QP and RQ of $triangle P Q R$ are produced to points S and T respectively. If $stack S P R with _ below equals 135 to the power of ring operator end exponent$ and $open floor P Q T equals 110 to the power of ring operator end exponent close$ then $stack P R Q with _ below equals$

In the given figure, if $P Q R equals 70 to the power of ring operator end exponent$ then $∣ A B C equals$