+19 Point D lies on side AC of equilateral triangle ABC such that the measure of angle DBC is 45 degrees. What is the ratio of the area of triangle ADB to the area of triangle CDB? Express your answer as a common fraction in simplest radical form.
+128473 Call the side of the equilateral triangle , S
Angle ABC = 60°
Angle DBC = 45°
And angle ABD = ABC - DBC = 60 - 45 = 15°
The area of triangle CDB =
(1/2) * S * BD * sin 45
And the area of triangle ADB =
(1/2) * S * BD * sin 15
So...the area of triangle ADB to the area of triangle CDB =
(1/2) * S * BD * sin 15 sin 15
_________________ = ______
(1/2 * S * BD * sin 45 sin 45
sin 15 = sin (45 - 30) =
sin45 cos30 - sin 30 cos 45 =
√2/2 * √3/2 - (1/2)(√2/2) =
[ √6 - √2 ] / 4
And sin 45 = √2/2
So sin 15 / sin 45 =
[ √6 - √2 ] / 4 * 2 / √2 =
[ √6 - √2 ] / [ 2√2 ] =
√6 / [ 2√2 ] - √2 / [ 2√2] =
√3 / 2 - 1/ 2 =
(√3 - 1 )
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