Question

ABCD is a rectangle with ad as longer side. P is a point such that APD forms an isosceles triangle with angle APD as 100. Diagonal AC is twice the length of side CD (smaller side of rectangle) diagonal AC interact PD at point X. What is angle AXP
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Answer 1

in the angle AXP is 50 deegree

Answer 2

Given :-

• ABCD is a rectangle .
• ∆APD is an isosceles triangle. .
• ∠APD = 100° .
• AC = 2CD
• Diagonal AC interact PD at point X.

To Find :-

• ∠AXP = ?

Construction :-

• Refer to image .

Solution :-

in isosceles triangle ∆APD , we have :-

→ ∠APD = 100° .

→ AP = PD .

So,

→ ∠PAD = ∠PDA = 40°. (Angle Opp. to Equal sides are Equal.)

Now, in ∆ADC :-

→ ∠ADC = 90° (Each angle of a rectangle is 90°)

→ AC = 2CD (Given)

→ AC/CD = (2/1)

So, By pythagoras theorem ,

→ AC² = CD² + AD²

→ (2)² = (1)² + AD²

→ AD² = 4 - 1

→ AD = √3

Than,

→ Perpendicular/Base = (1/√3)

→ Tan(∠DAC) = Tan30°

→ ∠DAC = 30° .

Therefore,

→ ∠PAX = ∠PAD - ∠DAC

→ ∠PAX = 40° - 30°

→ ∠PAX = 10°.

Hence, in ∆PAX

→ ∠PAX + ∠APX + ∠AXP = 180° (Angle sum Property).

→ 10° + 100° + ∠AXP = 180°

→ ∠AXP = 180° - 110°

→ ∠AXP = 70°(Ans.)