Question

In the given figure, PQ ⊥ AB, AQ = QB, ∠PAC = 42°, ∠C = 68°. Find ∠ABC.

Answer 1

Given:

△ABC

PQ ⊥ AB

AQ = QB Q is the midpoint of AB

∠PAC = 42°

∠C = 68°

Find ∠ABC

Step 1: Understand the figure

Q is the midpoint of AB

PQ ⊥ AB PQ is perpendicular from point P to AB

∠PAC = 42° Angle at point A

Step 2: Recall triangle angle sum

In △ABC: ∠ABC + ∠BCA + ∠BAC = 180°

We are given ∠C = 68°,

So, ∠ABC + ∠BAC + 68° = 180°

∠ABC + ∠BAC = 112°

Step 3: Use the given ∠PAC = 42°

∠PAC = 42°

Since AQ = QB, triangle △ABQ is isosceles.

∠PAC forms half of ∠BAC because Q is the midpoint of AB.

Thus, ∠BAC = 2 × ∠PAC = 2 × 42° = 84°

Step 4: Find ∠ABC

From Step 2: 

∠ABC + ∠BAC = 112°

Substitute ∠BAC = 84°:

∠ABC + 84 = 112°

∠ABC = 112 – 84°

∠ABC = 28°

Step 5: Reconsider the figure

AQ = QB Q is midpoint

PQ ⊥ AB PQ is altitude from P

PAC = 42° This is not necessarily half of BAC.

Instead, we can use the property of an isosceles triangle formed by the midpoint:

In triangle △ABQ, AQ = QB

PQ ⊥ AB PQ is perpendicular to the base of an isosceles triangle

Angle at vertex A is split by altitude PQ each part = 42°

So, angle BAC = angle PAC = 42° already given, not double.

Step 6: Apply triangle angle sum again

∠ABC + ∠BAC + ∠BCA = 180

∠ABC + 42 + 68 = 180

∠ABC + 110 = 180

∠ABC = 70

Step 7: Consider AQ = QB

Triangle △ABQ is isosceles with base AB

PQ ⊥ AB Triangle is right-angled at Q

Given ∠PAC = 42°, then by alternate interior angle property,

`∠ABC = 1/2 xx (180 - 68)`

∠ABC = 35°