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# help

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In triangle ABC, AB=3, AC=6, BC=8, and D lies on BC such that AD bisects $$\angle BAC$$. Find $$\cos \angle BAD$$

Oct 3, 2019
edited by Guest  Oct 3, 2019

#1
+9138
+1

In triangle ABC, AB= c = 3, AC= b = 6, BC= a = 8, and D lies on BC such that AD bisects $$\angle BAC$$. Find  $$\cos\angle BAC=cos( \alpha)$$

$$cosine:\\ a^2=b^2+c^2-2bc\cdot cos(\alpha)\\ cos(\alpha)= \frac{b^2+c^2-a^2}{2bc}=\frac{36+9-64}{2\cdot 6\cdot 3}$$

$$cos(\alpha)= \cos\angle BAC=-0,52\overline 7$$

!

Oct 3, 2019
#2
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Sorry asinus the question had a typo it should be "Find $$\cos \angle BAD$$"

Guest Oct 3, 2019
#3
+9138
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so:

$$cos(\alpha)= \cos\angle BAC=-0,52\overline 7$$

$$\angle BAD=\frac{1}{2} \angle BAC=\frac{1}{2}arccos (-0.52\overline 7)\\ \color{blue}\angle BAD= 60.928°$$

$$cos \angle BAD=0.48591$$

!

Oct 3, 2019
edited by asinus  Oct 3, 2019