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In ΔABC, ∠A : ∠B : ∠C = 3 : 3 : 4. A line parallel to BC is drawn which touches AB and AC at P and Q respectively. What is the value of ∠AQP - ∠APQ?
Options:
12
18
24
36
Let’s break it down:
- ?A : ?B : ?C = 3 : 3 : 4. That’s 3x, 3x, 4x. All angles in a triangle add up to 180°, so
3x + 3x + 4x = 180 ? 10x = 180 ? x = 18
So: ?A = ?B = 54°, ?C = 72°
- Now, a line parallel to BC cuts AB at P and AC at Q. PQ is parallel to BC, so triangle APQ is similar to triangle ABC.
- In these similar triangles, corresponding angles at A are equal. At P and Q, let’s label angles:
At Q: angle between AQ and PQ = angle QAP = a
At P: angle between PA and PQ = angle QPA = ß
- Since PQ is parallel to BC (let’s picture that), by similarity and alternate interior angles:
?AQP = ?B (so 54°)
?APQ = ?C (so 72°)
- So ?AQP - ?APQ = 54° - 72° = -18°
- Taking positive difference (as options are all positive): 18°
Here’s the answer:
- Option 1: 12 [nope, too low]
- Option 2: 18 This is right
- Option 3: 24 [doesn’t match]
- Option 4: 36 [way too high]
Option 2 is correct.
By: Kamal Kashyap ProfileResourcesReport error
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