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# Help Geometry

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I have found that angle C is 75 degrees, but I don't know what to do next.

Jun 10, 2024

#1
+257
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Triangle ACD is a 30-60-90 triangle as well, so AC = sqrt(3)⋅AD = sqrt(3)*24.

Since AD is an angle bisector in triangle ABC,

[\frac{BD}{CD} = \frac{AB}{AC} = \frac{48}{24 \sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}.]

Then CD=3/(2*sqrt(3)) = 6.

The area of triangle ABC is then 1/2⋅BC⋅AD = 1/2⋅48⋅24 = 576​.

Jun 10, 2024
#2
+118
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This answer happened to be wrong...:(

SilviaJendeukie  Jun 10, 2024
#3
+129733
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BAC  = 60

Angle ABC  =45

Therefore

DB / sin 30  = 24 / sin 45

DB  / (1/2)  = 24 /sin45

DB = (1/2) * 24 / [sqrt (2) / 2)   = 12sqrt 2

Angle ADB  =  180  - 45 - 30  =  105

So

AB / sin 105  = 24 / sqrt (2) /2

AB  = 24sin105 /[sqrt (2) /2 ]   = 12 + 12 sqrt (3)

AC / sin ABC  = AB /sin ACB

AC / (sqrt (2)  / 2) = (12 + 12 sqrt (3)) / sin 75

AC = (sqrt (2)  / 2) (12 + 12sqrt (3)) / sin 75   = 24

[ABC ] = (1/2) AB * AC * sin 60

[ABC ] = (1/2) (12 + 12 sqrt (3)) *24 * sqrt (3) /2 =  72 (3 + sqrt 3)  ≈ 340.7

Jun 10, 2024