In \(\triangle A B C, \angle A=\angle B=60^{\circ}, A C=\sqrt{ } 34 \mathrm{~cm}\). The lines \(A D\) and \(B D\) intersect at \(D\) with \(\angle D=\) \(90^{\circ}\). If \(\mathrm{DB}=\mathbf{3 \mathrm { cm }}\), then the length of \(\mathrm{AD}\) is:

Solution

Given:
– In \(\triangle ABC\), \(\angle A = \angle B = 60^\circ\), \(AC = \sqrt{34} \text{ cm}\).
– Lines \(AD\) and \(BD\) intersect at \(D\) with \(\angle D = 90^\circ\).
– \(DB = 3 \text{ cm}\).

Step 1: Determine the Type of Triangle

Since \(\angle A = \angle B = 60^\circ\), \(\angle C\) must also be \(60^\circ\) (as the sum of angles in a triangle is \(180^\circ\)). Therefore, \(\triangle ABC\) is an equilateral triangle.

Step 2: Find the Length of AB

Since \(\triangle ABC\) is equilateral, \(AB = AC = \sqrt{34} \text{ cm}\).

Step 3: Use Pythagoras’ Theorem

In \(\triangle ADB\), which is a right-angled triangle, we can use Pythagoras’ theorem:
\[ AB^2 = AD^2 + DB^2 \]
\[ (\sqrt{34})^2 = AD^2 + 3^2 \]
\[ 34 = AD^2 + 9 \]
\[ AD^2 = 34 – 9 \]
\[ AD^2 = 25 \]
\[ AD = \sqrt{25} \]
\[ AD = 5 \text{ cm} \]

Conclusion

The length of \(AD\) is 5 cm.