Fig. 15 The side BC of ∆ ABC is produced to a point D. the bisector of triangle A meets the side BC in P. If triangle ABC = 30° , triangle ACD = 110° , find APC .
Answers
Answer:
Step-by-step explanation:
Solution
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Given: AL is the bisector of ∠A, ∠ABC=30
∘
,∠ACD=115
∘
To find: ∠ALC
Solution: Clearly, ∠ACD+∠ACL=180
∘
[Linear pair]
⇒115
∘
∠ACL=180
∘
⇒∠ACL=180
∘
−115
∘
⇒∠ACL=65
∘
By Angle sum property of triangle in △ABC, we have
∠ABC+∠ACB+∠BAC=180
∘
⇒30
∘
+65
∘
+∠BAC=180
∘
⇒∠BAC=180
∘
−30
∘
−65
∘
⇒∠BAC=85
∘
Since AL is the bisector of ∠BAC
∴∠BAL=∠CAL=
2
∠BAC
=(
2
85
)
∘
=42.5
∘
Now, By angle sum property of triangle in △ALC, we have
∠ALC+∠CAL+∠ACL=180
∘
⇒∠ALC+42.5
∘
+65
∘
=180
∘
⇒∠ALC=180
∘
−42.5
∘
−65
∘
⇒∠ALC=72.5
∘
Answer:
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