For given parallelogram, the value of x = 114° and y = 66°
<h3>Explanation: </h3>
<em>(Refer figure)</em>
Given quadrilateral ABCD is a parallelogram.
m∠A = 66°
m∠B = x°
m∠C = y°
m∠D = 114°
Since by properties of parallelogram, the both pairs of opposite sides of a parallelogram are parallel.
Opposite angles of a parallelogram are congruent ...........................(1)
Consecutive angles of a parallelogram are supplementary ................(2)
Using (1) for given parallelogram ABCD; we get,
m∠B = m∠D
m∠A = m∠C
Therefore x = 114° and y = 66°.
Using (2) to check whether the two consecutive angles are supplementary.
( m∠A + m∠D ) = ( m∠B + m∠C ) = ( m∠C + m∠D ) = 66° + 114° = 180°
Do six multiply fifteen which equals to 90
I will use the letter x instead of theta.
Then the problem is, given sec(x) + tan(x) = P, show that
sin(x) = [P^2 - 1] / [P^2 + 1]
I am going to take a non regular path.
First, develop a little the left side of the first equation:
sec(x) + tan(x) = 1 / cos(x) + sin(x) / cos(x) = [1 + sin(x)] / cos(x)
and that is equal to P.
Second, develop the rigth side of the second equation:
[p^2 - 1] / [p^2 + 1] =
= [ { [1 + sin(x)] / cos(x) }^2 - 1] / [ { [1 + sin(x)] / cos(x)}^2 +1 ] =
= { [1 + sin(x)]^2 - [cos(x)]^2 } / { [1 + sin(x)]^2 + [cos(x)]^2 } =
= {1 + 2sin(x) + [sin(x)^2] - [cos(x)^2] } / {1 + 2sin(x) + [sin(x)^2] + [cos(x)^2] }
= {2sin(x) + [sin(x)]^2 + [sin(x)]^2 } / { 1 + 2 sin(x) + 1} =
= {2sin(x) + 2 [sin(x)]^2 } / {2 + 2sin(x)} = {2sin(x) ( 1 + sin(x)} / {2(1+sin(x)} =
= sin(x)
Then, working with the first equation, we have proved that [p^2 - 1] / [p^2 + 1] = sin(x), the second equation.