In a triangle ABC two points D,E are taken on BC so that angle BAD=angle DAE=angleCAE. Determine AE if AB=5,BC=10 angle BAC=90. PLEASE HELP I NEED HELP WITHIN TEN MINS PLEASE

Respuesta :

Answer:

AE = 7.5

Step-by-step explanation:

Since <BAC = [tex]90^{0}[/tex], then;

<BAD = <DAE = <CAE = [tex]30^{0}[/tex] (complementary angles)

From ΔABC, applying the Pythagoras theorem to determine the length of side AC;

[tex]/BC/^{2}[/tex] = [tex]/AC/^{2}[/tex] + [tex]/AB/^{2}[/tex]

[tex]/10/^{2}[/tex] = [tex]/AC/^{2}[/tex] + [tex]/5/^{2}[/tex]

100 = [tex]/AC/^{2}[/tex] + 25

[tex]/AC/^{2}[/tex] = 100 - 25

[tex]/AC/^{2}[/tex] = 75

AC = [tex]\sqrt{75}[/tex]

Applying trigonometric function to ΔCAE,

Cos [tex]30^{0}[/tex] = [tex]\frac{AE}{\sqrt{75} }[/tex]

AE = [tex]\sqrt{75}[/tex] × Cos [tex]30^{0}[/tex]

    = 7.5

Therefore, AE = 7.5

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