Create a quadratic polynomial, whose sum and product of zeroes are 1 and 3/2 respectively.

A quadratic polynomial with sum of zeroes $S$ and product of zeroes $P$ is given by the formula $x^2 - Sx + P$.
Given sum of zeroes $S = 1$.
Given product of zeroes $P = \frac{3}{2}$.
Substitute these values into the formula:

To remove the fraction, we can multiply the entire polynomial by 2:

The quadratic polynomial is:

The given quadratic equation is $2x^2 - 5x + 1 = 0$.
This is in the form $ax^2 + bx + c = 0$, where $a=2$, $b=-5$, and $c=1$.
The nature of the roots is determined by the discriminant $\Delta = b^2 - 4ac$.
Substitute the values of $a$, $b$, and $c$:

Since $\Delta > 0$, the roots are real and distinct.
The nature of the roots is:

The Basic Proportionality Theorem (also known as Thales Theorem) states that if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides proportionally.

The expression is $(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2$.
Expand each term using the identities $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$:

Now, add the two expanded expressions:

Combine like terms:

Factor out 2:

Using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$:

The value of the expression is:

The angle of elevation is the angle formed by the line of sight with the horizontal when an observer looks upwards at an object.

Let O be the center of the circle, A be the external point, and T be the point of tangency on the circle.
The length of the tangent from point A to the circle is $AT = 4$ cm.
The distance from point A to the center of the circle is $OA = 5$ cm.
The radius of the circle is $OT$.
We know that the radius is perpendicular to the tangent at the point of tangency, so $\triangle OTA$ is a right-angled triangle with the right angle at T.
By the Pythagorean theorem:

Substitute the given values:

Subtract $16\, \text{cm}^2$ from both sides:

Take the square root of both sides:

The radius of the circle is: