In triangle ABC, AB = AC = 15 cm and BC = 18 cm, find cos `angleABC`

Answer

To find \( \cos \angle ABC \) in triangle ABC where \( AB = AC = 15 \, \text{cm} \) and \( BC = 18 \, \text{cm} \), we can follow these steps:

### Step 1: Identify the triangle type
Since \( AB = AC \), triangle ABC is an isosceles triangle.

### Step 2: Draw a perpendicular from A to BC
To simplify calculations, we can draw a perpendicular line from point A to line segment BC. Let’s denote the foot of the perpendicular as point D. This will help us create two right triangles, ABD and ACD.

### Step 3: Calculate the length of BD and DC
Since AD is a perpendicular bisector in an isosceles triangle, it divides BC into two equal segments. Therefore, we can find:
\[ BD = DC = \frac{BC}{2} = \frac{18}{2} = 9 \, \text{cm} \]

### Step 4: Use the Pythagorean theorem in triangle ABD
In triangle ABD, we can apply the Pythagorean theorem:
\[ AB^2 = AD^2 + BD^2 \]
Substituting the known values:
\[ 15^2 = AD^2 + 9^2 \]
\[ 225 = AD^2 + 81 \]
\[ AD^2 = 225 - 81 = 144 \]
\[ AD = \sqrt{144} = 12 \, \text{cm} \]

### Step 5: Calculate \( \cos \angle ABC \)
In triangle ABD, we know:
- The base (BD) = 9 cm
- The hypotenuse (AB) = 15 cm

Using the definition of cosine:
\[ \cos \angle ABC = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BD}{AB} = \frac{9}{15} = \frac{3}{5} \]

### Final Answer
Thus, \( \cos \angle ABC = \frac{3}{5} \).

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