In the right △ABC, CD is the altitude to the hypotenuse AB, and m∠ABC=30


. Find AD, if BD=54 cm.

1 answer

In right triangle \( ABC \), with \( \angle ABC = 30^\circ \), we can set up our relationships and calculations using trigonometric ratios and the properties of similar triangles.

We know that in triangle \( ABC \):

  • \( AB \) is the hypotenuse,
  • \( BC \) is opposite the angle \( 30^\circ \),
  • \( AC \) is adjacent to the angle \( 30^\circ \),
  • \( D \) is the foot of the altitude from \( C \) to \( AB \),
  • \( BD = 54 , cm \).

Step 1: Identify the lengths in triangle \( ABC \)

Since \( \angle ABC = 30^\circ \), we can identify the ratios of the sides relative to angle \( B \):

  • The side opposite \( 30^\circ \) (which is side \( AC \)) is half the length of the hypotenuse \( AB \): \[ AC = \frac{1}{2} AB \]
  • Let \( AB = c \), then \( AC = \frac{1}{2} c \).

Step 2: Use the Pythagorean theorem in triangle \( ABC \)

Let \( BC = a \). The right triangle \( ABC \) satisfies: \[ AB^2 = AC^2 + BC^2 \] Substituting the values we have: \[ c^2 = \left( \frac{1}{2} c \right)^2 + a^2 \] \[ c^2 = \frac{1}{4} c^2 + a^2 \] Rearranging gives: \[ c^2 - \frac{1}{4} c^2 = a^2 \] \[ \frac{3}{4} c^2 = a^2 \quad \Rightarrow \quad a = \frac{\sqrt{3}}{2} c \]

Step 3: Use the relationship between AD, BD, and hypotenuse

In right triangles formed by dropping the altitude \( CD \):

  1. From triangle \( BCD \): \[ CD = a \sin 30^\circ = a \cdot \frac{1}{2} = \frac{1}{2} a \]

  2. From triangle \( ACD \): \[ CD = b \sin 30^\circ = b \cdot \frac{1}{2} = \frac{1}{2} b \]

where \( b \) is the length \( AD \) (the side opposite angle \( A \)).

By the similarity of triangles \( ABC \) and \( BCD \) (and \( ACD \)), we also know: \[ \frac{AD}{AC} = \frac{BD}{BC} \quad \Rightarrow \quad \frac{AD}{\frac{1}{2}c} = \frac{54}{\frac{\sqrt{3}}{2}c} \]

Now, this simplifies to: \[ \frac{AD}{\frac{1}{2}c} = \frac{54 \cdot 2}{\sqrt{3} c} = \frac{108}{\sqrt{3} c} \] Cross-multiplying yields: \[ AD \cdot \sqrt{3} c = 54c \quad \Rightarrow \quad AD = \frac{54}{\sqrt{3}} = 18\sqrt{3} \approx 31.18 , cm \]

Conclusion

Thus, the length of \( AD \) is: \[ \boxed{18\sqrt{3}} , cm. \]