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Right triangle ABC has hypotenuse AC, angle CAB=30°, and BC=√2. Right triangle ACD has hypotenuse AD and angle DAC=45°. The interiors of ABC and ACD do not overlap. Find the length of the perpendicular from D onto AB.

Can someone please provide a solution to this problem? Any help is greatly appreciated.

Michael Joyce
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Lulu Uy
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1 Answers1

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We have $$sin 30^o = \frac{1}{2} = \frac{BC}{AC} = \frac{\sqrt{2}}{AC}$$

so $AC = 2\sqrt2$.

Similarly $$cos 45^o = \frac{1}{\sqrt2} = \frac{AC}{AD} = \frac{2\sqrt2}{AD}$$

so $AD = 4$.

Finally, let $D_p$ denote described perpendicular. Then $$ sin 75^o = \frac{D_p}{AD}$$

so $D_p = 4 \times sin 75^o$.

gamma
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  • I understand how you found AD and AC, but I'm still not quite sure how to find the perpendicular from D onto AB. Also this a multiple choice practice test, so the answer choices are: A. √6+2√3 B. √6+√2 C. 4 D. 1+√5 (thank you for your help) – Lulu Uy Oct 22 '14 at 22:33
  • @EdElric sin 75 = $ \frac{\sqrt{6} + \sqrt{2}}{4} $ so the answer would be B. The 75 comes form $ CAB = 30^o $ plus $ DAC = 45^o $ . – gamma Oct 22 '14 at 23:07