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1) I is the incenter of triangle ABC. If angle ABC = 78^\circ$ and angle ACB = 47^\circ,$ find angle BID, in degrees.

2) Find angle C in degrees.


3) I is the incenter of triangle ABC. We have CE = 15, AE = 6, and AB = 10. Find AI/ID provided that BF=125/23, and BC=25.


4) The incircle of triangle ABC is shown. Find the area of triangle ABC


5) Find the area of triangle ACE


THANK YOU SO MUCH!!!!!! It would be great if these could be solved by next week!

 Apr 21, 2024
 #1
avatar+1859 
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Problem 1)

 

Since ∠ABC=78∘ and ∠ACB=47∘, then ∠BAC=180∘−78∘−47∘=55∘.

 

Let D be the foot of the altitude from B to AC. Then ∠ABD=∠ABC=78∘.

 

Triangle ABD is a 30-60-90 triangle, so ∠ADB=60∘ and ∠BAD=30∘.

 

Since I is the incenter, ∠BAI=∠CAI=∠BAC/2=27.5∘. Then ∠ABI=∠BAI+∠BAD=57.5∘.

 

Triangle ABI is a 30-57.5-92.5$ triangle, so ∠BID = ∠ABI/2 = 46.25∘​.

 Apr 24, 2024
 #2
avatar+1859 
+1

Problem 2)

 

Since BD=AD=AC then triangle ABC is isosceles with AB=AC. Therefore, ∠ABC=∠ACB=x (for some angle measure x ).

 

Since D is on side BC so BD=AD, then line AD is an angle bisector of ∠A. Therefore, ∠BAD=∠CAD=2∠A​.

 

Since the angles in a triangle sum to 180∘, then in triangle ABC, ∠A+∠B+∠C=180∘. Substituting x for ∠B and ∠C, we get ∠A+x+x=180∘.

 

Solving for ∠A, we get ∠A=180∘−2x.

 

We are also given that ∠BAD=∠CAD=2∠A​. Substituting the expression for ∠A that we just found, we get 2180∘−2x​=∠BAD=∠CAD.

 

Because the sum of the angles in a triangle is 180∘, then in triangle ABD, ∠BAD+∠ABD+∠ADB=180∘.

 

Substituting 78∘ for ∠ABD (since BD=AD=AC and triangle ABC is isosceles), we get 2180∘−2x​+78∘+90∘=180∘.

 

Solving for x, we get x=31∘​.

 

Therefore, angle ACB = 31 degrees.

 Apr 24, 2024

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