1. The problem involves a right triangle ABC with right angle at C, hypotenuse AB, and altitude CD drawn perpendicular to AB at D. 2. We are given three parts: a) Given AB = 24, find AD. b) Given AB = 20 and \(\angle A = 60^\circ\), find side a (which is BC). c) Given AC = 10, AD = 4, find BD. 3. Important formulas and rules: - In a right triangle, the altitude to the hypotenuse creates two smaller right triangles similar to the original. - The altitude length \(CD\) satisfies \(CD^2 = AD \times BD\). - The segments on the hypotenuse satisfy \(AB = AD + BD\). - Using trigonometry, \(\sin(\theta) = \frac{opposite}{hypotenuse}\). 4. Part a) Given \(AB = 24\) and \(\angle B = 30^\circ\) (from diagram), find AD. - Since \(\angle B = 30^\circ\), side AC opposite to \(30^\circ\) is half the hypotenuse: \(AC = \frac{AB}{2} = 12\). - Using Pythagoras, \(BC = \sqrt{AB^2 - AC^2} = \sqrt{24^2 - 12^2} = \sqrt{576 - 144} = \sqrt{432} = 12\sqrt{3}\). - The altitude divides the hypotenuse into segments AD and BD. - Using similarity, \(AD = AC^2 / AB = \frac{12^2}{24} = \frac{144}{24} = 6\). 5. Part b) Given \(AB = 20\) and \(\angle A = 60^\circ\), find side a (BC). - Using sine, \(\sin 60^\circ = \frac{BC}{AB} \Rightarrow BC = AB \times \sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.32\). - Using cosine, \(AC = AB \times \cos 60^\circ = 20 \times \frac{1}{2} = 10\). 6. Part c) Given \(AC = 10\), \(AD = 4\), find BD. - Since \(AB = AD + BD\), and \(AB^2 = AC^2 + BC^2\), we first find \(AB\). - Using similarity, \(AD = AC^2 / AB \Rightarrow 4 = \frac{10^2}{AB} = \frac{100}{AB} \Rightarrow AB = \frac{100}{4} = 25\). - Then, \(BD = AB - AD = 25 - 4 = 21\). Final answers: - a) \(AD = 6\) - b) \(BC = 10\sqrt{3} \approx 17.32\) - c) \(BD = 21\)