Question

The angles of a quadrilateral are $\left( {3x + 2} \right)^\circ $,$\left( {x - 3} \right)^\circ $, $\left( {2x + 1} \right)^\circ $, $2\left( {2x + 5} \right)^\circ $respectively. Find the value of x and the measure of each angle.

Answer 1

Hint: We can find the sum of the angles of a quadrilateral using the equation $\left( {n - 2} \right) \times 180^\circ $where n is the number of sides. Then we can find the sum of the angles by adding the given angles. Equating the sum of angles, we get a solution in x. We can find the value of x by solving the equation. We can find the measure of each angle by substituting x in each of the given angles.

Complete step by step Answer:

We know that a quadrilateral has 4 sides.
$ \Rightarrow n = 4$
We can find the sum of the angles of a quadrilateral using the equation,
$S = \left( {n - 2} \right) \times 180^\circ $
On substituting the value of n, we get,
$ \Rightarrow S = \left( {4 - 2} \right) \times 180^\circ $
On further calculation, we get,
$ \Rightarrow S = 360^\circ $.. (1)
The angles are given as $\left( {3x + 2} \right)^\circ $,$\left( {x - 3} \right)^\circ $, $\left( {2x + 1} \right)^\circ $, $2\left( {2x + 5} \right)^\circ $.
We can take their sum,
\[ \Rightarrow S = \left( {3x + 2} \right)^\circ + \left( {x - 3} \right)^\circ + \left( {2x + 1} \right)^\circ + 2\left( {2x + 5} \right)^\circ \]
On simplification, we get,
\[ \Rightarrow S = \left( {10x + 10} \right)^\circ \].. (2)
As (1) and (2) are sum of the angles of a quadrilateral, we can equate them.
\[ \Rightarrow \left( {10x + 10} \right)^\circ = 360^\circ \]
On subtracting 10 on both sides, we get,
\[ \Rightarrow 10x^\circ = 350^\circ \]
On diving throughout with 10, we get,
\[ \Rightarrow x = 35\]
Therefore, the value of x is 35.
To find the measure of the angles we substitute the value of x
$ \Rightarrow \left( {3x + 2} \right)^\circ = \left( {3 \times 35 + 2} \right)^\circ $
$ = \left( {105 + 2} \right)^\circ $
$ = 107^\circ $
For the second angle,
$ \Rightarrow \left( {x - 3} \right)^\circ = \left( {35 - 3} \right)^\circ $
$ = 32^\circ $
For 3rd angle,
$ \Rightarrow \left( {2x + 1} \right)^\circ = \left( {2 \times 35 + 1} \right)^\circ $
$ = \left( {70 + 1} \right)^\circ $
$ = 71^\circ $
For the 4th angle,
$ \Rightarrow 2\left( {2x + 5} \right)^\circ = 2\left( {2 \times 35 + 5} \right)^\circ $
$ = 2\left( {70 + 5} \right)^\circ $
$ = 2\left( {75} \right)^\circ $
$ = 150^\circ $
Therefore the angles are $107^\circ $,$32^\circ $,$71^\circ $,$150^\circ $

Note: A quadrilateral is a polygon with 4 sides and 4 angles. Sum of the interior angles of a quadrilateral is $360^\circ $. The sum of the interior angles of polygons with n sides are given by the equation $\left( {n - 2} \right) \times 180^\circ $. In this problem, all the angles were given as linear expressions with x as a parameter. So, we add all the angles and equated to the angle sum to form a linear equation on x. As there is only one variable, we can solve it using the same equation. We cannot solve for variables if the number of equations is less than the number of variables.