Question

in the given figure, lones ABa nd CD intersect at a point P such that ∠PAC=45°, ∠ACP=100° and ∠PBD=65°, Find ∠CPA,∠DPB and ∠BDP.

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Answer 1

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$\huge{\underline{\underline{\sf{\orange{</p><p>Question:-}}}}}$

In the given figure, lones ABa nd CD intersect at a point P such that ∠PAC=45°, ∠ACP=100° and ∠PBD=65°, Find ∠CPA,∠DPB and ∠BDP.

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We know that the sum of the angles of a triangle is 180°.

In ∆ACP,we have

$\angle \: PAC + \angle \: ACP + \angle \: CPA = 180\degree$

$⇒ 45\degree + 100\degree + \angle \: CPA = 180\degree$

$⇒ \angle \: CPA = (180\degree - 145\degree) = 35\degree$

$\therefore \: \angle \: DPB = \angle \: CPA = 35\degree$

In ∆PBD, we have

$\angle \: DPB +\angle \: PBD + \angle \: BDP = 180\degree \\ \: \: \: \: [sum \: of \: the \: angles \: of \: a \: triangle]$

$⇒ 35\degree + 65\degree + \angle \: BDP = 180\degree$

$⇒ 100\degree + \angle \: BDP = 180\degree$

$⇒ \angle \: BDP = (180\degree - 100\degree) = 80\degree.$

Hence,∠CPA = 35°, ∠DPB=35° and ∠BDP=80°

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Answer 2

Answer:

Angle APC = 35°

Angle DPB = 35°

Angle PDB = 80°

Step-by-step explanation:

• as we know that the sum of interior angles of a triangle are always 180°

so the angles of ∆APC= 45+ 100+ x

180-(45+100) = 35°

so the measure of angle apc is 35°

• if angle apc is 35° then the opposite angle of angle apc is also 35°

means angle dpb( which is opposite to angle apc) is also 35°

• so now we just have to find the angle pdb= 35°+ 65° + x

= 180-(65+35) = 80

so the angle pdb is 80°

• Angle APC = 35°
• Angle DPB = 35°
• Angle PDB = 80°