Question

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

Answer 1

Answer:

AE = 7.5

Step-by-step explanation:

Since <BAC = $90^{0}$, then;

<BAD = <DAE = <CAE = $30^{0}$ (complementary angles)

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

$/BC/^{2}$ = $/AC/^{2}$ + $/AB/^{2}$

$/10/^{2}$ = $/AC/^{2}$ + $/5/^{2}$

100 = $/AC/^{2}$ + 25

$/AC/^{2}$ = 100 - 25

$/AC/^{2}$ = 75

AC = $\sqrt{75}$

Applying trigonometric function to ΔCAE,

Cos $30^{0}$ = $\frac{AE}{\sqrt{75} }$

AE = $\sqrt{75}$ × Cos $30^{0}$

    = 7.5

Therefore, AE = 7.5