Question

In ΔABC, the perpendiculars drawn from A,B and C meet the opposite sides at D,E and F, respectively.AD, BE and CF intersect at point P. If ∠EPD=116⁰ and the bisectors of ∠A and ∠B meet at Q, then the measure of ∠AQB?

Plz give proper explanations...

Answer 1

Given :- (from image.)

• In ΔABC, the perpendiculars drawn from A,B and C meet the opposite sides at D,E and F, respectively.
• AD, BE and CF intersect at point P.
• ∠EPD=116° .
• The bisectors of ∠A and ∠B meet at Q .

To Find :-

• ∠AQB = ?

Solution :-

As we know angle bisectors meets at incentre .

So,

• Q = incentre of ∆ABC .

then,

• ∠AQB = 90° + (C/2) .

Now, in quadrilateral EPDC, we have,

• ∠PEC = 90°
• ∠PDC = 90°
• ∠EPD = 116°

then, by angle sum property ,

→ ∠PEC + ∠PDC + ∠EPD + ∠C = 360°

→ 90° + 90° + 116° + ∠C = 360°

→ ∠C = 360° - 296°

→ ∠C = 64° .

Therefore,

→ ∠AQB = 90° + (C/2)

→ ∠AQB = 90° + (64/2)

→ ∠AQB = 90° + 32°

→ ∠AQB = 122° (Ans.)

Hence, ∠AQB is equal to 122°.

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Answer 2

Refer the attachment ↑

∠AQB = 112°