Question

In ∆ABC and ∆ POR,angle A = 30°, AngleB = 70°, angle P = 70°, angle Q = 80° and AB = RP. Write the correspondence if triangles are congruent​

Answer 1

Answer:

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Answer 2

$\red{\bf{\bigstar \: \: {\underline{\underline{ Given :}}}}}$

• ∆ ABC ≅ ∆ POR
• ∠A = 30°
• ∠B = 70°
• ∠P = 70°
• ∠Q = 80°
• AB = RP

$\red{\Large{\bf{\bigstar \: \:{\underline{\underline{ Solution :}}}}}}$

Here we have to find all corresponding parts & angles.

■ In ∆ ABC

• ∠A = 30°
• ∠B = 70°
• ∠C = ?°

So, ∠C is :-

⟹ ∠A + ∠B + ∠C = 180° [ By angle sum property of triangle ]

⟹ 30° + 70° + ∠C = 180°

⟹ ∠C + 100° = 180°

⟹ ∠C = 180° - 100

⟹ ∠C = 80°

So, ∠C is 80°.

■ In ∆ PQR

• ∠P = 70°
• ∠Q = 80°
• ∠R = ?°

So, ∠R is :-

⟹ ∠P + ∠Q + ∠R = 180° [ By angle sum property of triangle ]

⟹ 70° + 80° + ∠R = 180°

⟹ ∠R + 150° = 180°

⟹ ∠R = 180° - 150°

⟹ ∠R = 30°

So, ∠R is 30°.

So, here corresponding angles are :-

• ∠A = ∠R [30°]
• ∠B = ∠P [70°]
• ∠C = ∠Q [80°]

Here corresponding sides are :-

• AB = RP
• BC = PQ
• AC = QR

● Rough diagram of figure :-

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A }\put(0.5,-0.3){ \bf B }\put(5.2,-0.3){ \bf C }\end{picture}$ $\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf P }\put(0.5,-0.3){ \bf Q }\put(5.2,-0.3){ \bf R }\end{picture}$