Identifying the Quadrilateral

The quadrilateral $ABCD$ has diagonals $AC$ and $BD$ that are perpendicular, and the lengths of sides $AB = 10$ and $AD = 10$, with $\angle A = 100^\circ$ and $\angle D = 64^\circ$. This suggests that $ABCD$ is a kite, where two pairs of adjacent sides are equal in length (i.e., $AB = AD$ and $AB$ is adjacent to $AD$).

Step-by-Step Solution

We need to use the properties of a kite and trigonometry to solve for the lengths of the diagonals $AC$ and $BD$.

Step 1: Key Geometry of the Kite

• Kite Properties:
  • Diagonals intersect at right angles (perpendicular), as given.
  • The diagonal connecting the pair of equal sides bisects the other diagonal at right angles.

Since $AB = AD$, the diagonal $BD$ will be bisected by diagonal $AC$ at point $E$.

Step 2: Triangle Analysis

We will use trigonometry in triangles $\triangle ABD$ and $\triangle ADB$ to solve for the diagonals.

Step 3: Using Law of Cosines in Triangle $ABD$

Since $AB = AD = 10$, we can use the law of cosines to find the length of diagonal $BD$.

$\text{In } \triangle ABD: \quad BD^2 = AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos(\angle A)$ Substitute the given values:

$BD^2 = 10^2 + 10^2 - 2 \cdot 10 \cdot 10 \cdot \cos(100^\circ)$ $BD^2 = 100 + 100 - 200 \cdot \cos(100^\circ)$ Using $\cos(100^\circ) \approx -0.1736$:

$BD^2 = 200 - 200 \cdot (-0.1736)$ $BD^2 = 200 + 34.72$ $BD^2 = 234.72$ $BD \approx \sqrt{234.72} \approx 15.3$

Thus, the length of diagonal $BD$ is approximately 15.3 units.

Step 4: Using Sine Rule in Triangle $ABD$ for Diagonal $AC$

Now, we use the sine rule to find diagonal $AC$. We first need the angle $\angle ABD$, which can be found by recognizing that the sum of the angles in triangle $ABD$ is $180^\circ$. Given that $\angle A = 100^\circ$ and $\angle D = 64^\circ$, we calculate $\angle ABD$:

$\angle ABD = 180^\circ - \angle A - \angle D = 180^\circ - 100^\circ - 64^\circ = 16^\circ$

Using the sine rule in triangle $ABD$:

$\frac{AC}{\sin(16^\circ)} = \frac{BD}{\sin(100^\circ)}$

Substitute the known values:

$\frac{AC}{\sin(16^\circ)} = \frac{15.3}{\sin(100^\circ)}$ Since $\sin(100^\circ) \approx \sin(80^\circ) \approx 0.9848$, we have:

$\frac{AC}{\sin(16^\circ)} = \frac{15.3}{0.9848}$ $\frac{AC}{0.2756} = 15.5$ $AC \approx 15.5 \times 0.2756 \approx 4.27$

Thus, the length of diagonal $AC$ is approximately 4.27 units.

Final Answer:

• Length of diagonal $AC \approx 4.27$ units
• Length of diagonal $BD \approx 15.3$ units

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Would you like further details on the steps or the trigonometric calculations? Or do you have any follow-up questions?

Here are some related questions you might find interesting:

1. How does the law of cosines apply in other non-right triangles?
2. What is the relationship between the angles in a kite-shaped quadrilateral?
3. How does the sine rule work, and in which types of triangles is it useful?
4. Can we find other properties of a kite using its diagonals?
5. What is the importance of bisecting diagonals in solving for unknown side lengths?

Tip: Always remember that in kite-shaped quadrilaterals, the diagonals are perpendicular and the one connecting the two pairs of equal sides bisects the other diagonal.