In a `DeltaABC, B=90^(@), AC=h` and the length of perpendicular from B to AC is p such that `h=4p.` If `AB lt BC,` then measure `angleC.`

Answer

To solve the problem step by step, we will use the properties of right triangles and trigonometric identities.

### Step 1: Understand the Triangle Configuration
We are given a right triangle \( \Delta ABC \) where \( \angle B = 90^\circ \), \( AC = h \), and the length of the perpendicular from \( B \) to \( AC \) is \( p \). We also know that \( h = 4p \).

### Step 2: Set Up the Triangle
Let \( D \) be the foot of the perpendicular from \( B \) to \( AC \). Thus, \( BD = p \) and \( AD + DC = AC = h = 4p \).

### Step 3: Define the Segments
Let \( AD = x \) and \( DC = 4p - x \). Since \( AB < BC \), we can assume \( C \) is at the right side of \( B \) and \( A \) is at the left side.

### Step 4: Use Trigonometric Ratios
In triangle \( BDC \):
\[ \tan C = \frac{BD}{DC} = \frac{p}{4p - x} \]

In triangle \( BDA \):
\[ \tan A = \frac{BD}{AD} = \frac{p}{x} \]

### Step 5: Relate the Tangents
From the definitions of tangent:
\[ \tan C = \frac{p}{4p - x} \quad \text{and} \quad \tan A = \frac{p}{x} \]

### Step 6: Set Up the Equation
Using the property of tangent:
\[ \tan A \cdot \tan C = 1 \]
Substituting the expressions we have:
\[ \frac{p}{x} \cdot \frac{p}{4p - x} = 1 \]
This simplifies to:
\[ \frac{p^2}{x(4p - x)} = 1 \]
Thus:
\[ p^2 = x(4p - x) \]
Rearranging gives:
\[ x^2 - 4px + p^2 = 0 \]

### Step 7: Solve for \( x \)
Using the quadratic formula:
\[ x = \frac{4p \pm \sqrt{(4p)^2 - 4 \cdot 1 \cdot p^2}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{4p \pm \sqrt{16p^2 - 4p^2}}{2} = \frac{4p \pm \sqrt{12p^2}}{2} = \frac{4p \pm 2\sqrt{3}p}{2} = 2p \pm \sqrt{3}p \]
Thus:
\[ x = (2 \pm \sqrt{3})p \]

### Step 8: Find \( DC \)
Using \( DC = 4p - x \):
\[ DC = 4p - (2 \pm \sqrt{3})p = (2 \mp \sqrt{3})p \]

### Step 9: Calculate \( \tan C \)
Now we can find \( \tan C \):
\[ \tan C = \frac{p}{DC} = \frac{p}{(2 \mp \sqrt{3})p} = \frac{1}{2 \mp \sqrt{3}} \]

### Step 10: Find \( \angle C \)
Using the inverse tangent function:
\[ \angle C = \tan^{-1}\left(\frac{1}{2 \mp \sqrt{3}}\right) \]

### Conclusion
To find the exact angle, we can evaluate the expression. However, we can also recognize that:
\[ \angle C = 15^\circ \quad \text{(since \( \tan 15^\circ = 2 - \sqrt{3} \))} \]

Thus, the measure of angle \( C \) is \( 15^\circ \).

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