Quadrilaterals - Angles in a Quadrilateral

Three angles of a quadrilateral are $75^\circ$, $90^\circ$, and $75^\circ$. Find the fourth angle.

Let the fourth angle be $x$. According to the angle sum property: $75^\circ + 90^\circ + 75^\circ + x = 360^\circ$ First, add the given angles: $\begin{array}{r} 75 \\ 90 \\ + 75 \\ \hline 240 \end{array}$ So, $240^\circ + x = 360^\circ$ $x = 360^\circ - 240^\circ$ $\begin{array}{r} 360 \\ - 240 \\ \hline 120 \end{array}$ Therefore, $x = 120^\circ$.

We use the Angle Sum Property which states that all interior angles must add up to $360^\circ$. By subtracting the sum of the three known angles from $360^\circ$, we find the unknown angle.

The angles of a quadrilateral are in the ratio $3:5:9:13$. Find all the angles of the quadrilateral.

Let the common ratio multiplier be $x$. Therefore, the angles are $3x$, $5x$, $9x$, and $13x$. By the angle sum property: $3x + 5x + 9x + 13x = 360^\circ$ $30x = 360^\circ$ $x = \frac{360^\circ}{30} = 12^\circ$ Now, calculate each angle: $3x = 3 \times 12^\circ = 36^\circ$ $5x = 5 \times 12^\circ = 60^\circ$ $9x = 9 \times 12^\circ = 108^\circ$ $13x = 13 \times 12^\circ = 156^\circ$

When angles are given in a ratio, we represent them as terms of $x$, sum them to $360^\circ$, solve for $x$, and then substitute $x$ back into each term to find the individual angle measures.

In a quadrilateral $ABCD$, $\angle A = (x + 10)^\circ$, $\angle B = (2x + 5)^\circ$, $\angle C = (x - 15)^\circ$, and $\angle D = (x + 20)^\circ$. Find the value of $x$.

Sum of angles $= 360^\circ$ $(x + 10) + (2x + 5) + (x - 15) + (x + 20) = 360$ Grouping like terms: $(x + 2x + x + x) + (10 + 5 - 15 + 20) = 360$ $5x + 20 = 360$ $5x = 360 - 20$ $5x = 340$ $x = \frac{340}{5}$ $x = 68$

We set up an algebraic equation based on the Angle Sum Property. By combining the coefficients of $x$ and the constant terms, we solve for the variable $x$.

1. Find the interior angles using the linear pair property: Interior $\angle P = 180^\circ - 110^\circ = 70^\circ$ Interior $\angle Q = 180^\circ - 80^\circ = 100^\circ$ Interior $\angle R = 180^\circ - 120^\circ = 60^\circ$
2. Let the fourth interior angle be $x$.
3. Using Angle Sum Property: $70^\circ + 100^\circ + 60^\circ + x = 360^\circ$ $230^\circ + x = 360^\circ$ $x = 360^\circ - 230^\circ = 130^\circ$

Therefore, the value of $x$ is $130^\circ$.

We first convert the given exterior angles to interior angles because the sum of interior angles of any quadrilateral is always $360^\circ$.

1. Since $AB \parallel DC$, the adjacent angles between the parallel lines (consecutive interior angles) are supplementary.
2. $\angle A + \angle D$ does not apply here as they are on the same transversal $AD$. $\angle A + \angle D = 55^\circ + 70^\circ = 125^\circ$.
3. For $AB \parallel DC$, $\angle A + \angle D$ are interior angles on the same side of transversal $AD$ only if $AD$ was the parallel pair. Here $AB \parallel DC$, so $\angle A + \angle D$ are not necessarily $180^\circ$ unless it's a specific shape.
4. However, in a trapezoid where $AB \parallel DC$, $\angle A + \angle D = 180^\circ$ and $\angle B + \angle C = 180^\circ$ is incorrect. The correct relation is $\angle A + \angle D = 180^\circ$ (if $AD$ is transversal) and $\angle B + \angle C = 180^\circ$.
5. Given $\angle A = 55^\circ$, then $\angle D = 180^\circ - 55^\circ = 125^\circ$. But the problem states $\angle D = 70^\circ$. This implies the parallel sides are $AD$ and $BC$.
6. If $AD \parallel BC$, then $\angle A + \angle B = 180^\circ \implies \angle B = 180^\circ - 55^\circ = 125^\circ$.
7. Also $\angle D + \angle C = 180^\circ \implies \angle C = 180^\circ - 70^\circ = 110^\circ$.
8. Check: $55 + 125 + 110 + 70 = 360^\circ$.

When two sides of a quadrilateral are parallel, the angles interior to the parallel lines on the same transversal sum to $180^\circ$.