Question

solve the following problems
Given:Triangle AKL,AK=9,m<K=90 degrees, m<A=60 degrees.
Find:The area of the triangle.

Answer 1

To find the area of triangle AKL where \( \angle K = 90^\circ \), \( AK = 9 \), and \( \angle A = 60^\circ \), we can first determine the lengths of the other sides using trigonometric ratios.

Since \( \angle K \) is the right angle, we can use properties of a right triangle. Let's denote the lengths of the sides:
- Side opposite to angle \( A \) (which is \( KL \))
- Side opposite to angle \( L \) (which is \( AL \))

### Step 1: Determine the Length of Side \( AL \)

Using the sine function:
\[
\sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{KL}{AK}
\]
\[
\sin(60^\circ) = \frac{KL}{9}
\]

Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[
\frac{\sqrt{3}}{2} = \frac{KL}{9}
\]

Multiplying both sides by 9:
\[
KL = 9 \cdot \frac{\sqrt{3}}{2} = \frac{9\sqrt{3}}{2}
\]

### Step 2: Determine the Length of Side \( AC \) (which is \( AL \))

Using the cosine function:
\[
\cos(A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AL}{AK}
\]
\[
\cos(60^\circ) = \frac{AL}{9}
\]

Since \( \cos(60^\circ) = \frac{1}{2} \):
\[
\frac{1}{2} = \frac{AL}{9}
\]

Multiplying both sides by 9:
\[
AL = 9 \cdot \frac{1}{2} = \frac{9}{2}
\]

### Step 3: Calculate the Area of Triangle AKL

The area \( A \) of a right triangle can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]

Using \( AL \) and \( KL \) as the height and base respectively:
\[
\text{Area} = \frac{1}{2} \times AL \times KL
\]
\[
= \frac{1}{2} \times \frac{9}{2} \times \frac{9\sqrt{3}}{2}
\]
\[
= \frac{1}{2} \times \frac{81\sqrt{3}}{4}
\]
\[
= \frac{81\sqrt{3}}{8}
\]

Thus, the area of triangle AKL is:
\[
\text{Area} = \frac{81\sqrt{3}}{8} \, \text{square units}
\]