Plane Geometry
-  In a ∆ABC, if 2 ∠ A = 3 ∠ B = 6∠C, value of ∠ B is
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                        View Hint View Answer Discuss in Forum As per the given in question, 
 ∴ 2∠A = 3∠B = 6∠C⇒ 2∠A = 3∠B = 6∠C 6 6 6 ⇒ ∠A = ∠B = ∠C 3 2 1 
 ⇒ ∠A : ∠B : ∠C = 3 : 2 : 1∴ ∠B =  2  × 180° 1 + 2 + 3 
 Correct Option: AAs per the given in question, 
 ∴ 2∠A = 3∠B = 6∠C⇒ 2∠A = 3∠B = 6∠C 6 6 6 ⇒ ∠A = ∠B = ∠C 3 2 1 
 ⇒ ∠A : ∠B : ∠C = 3 : 2 : 1∴ ∠B =  2  × 180° 1 + 2 + 3 ∠B = 2 × 180° = 60° 6 
-  In a ∆ABC, ∠A + ∠B = 70° and ∠B + ∠C = 130°, value of ∠A is
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                        View Hint View Answer Discuss in Forum In ∆ ABC, 
 As we know that , ∠A + ∠B + ∠C = 180°
 Given , ∠A + ∠B = 70° and ∠B + ∠C = 130°Correct Option: BIn ∆ ABC, 
 As we know that , ∠A + ∠B + ∠C = 180°
 Given , ∠A + ∠B = 70° and ∠B + ∠C = 130°
 ∴ ∠A = (∠A + ∠B + ∠C) – (∠B + ∠C)
 ∠A = 180– 130°= 50°
-  Angle between sss ∠B is
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                        View Hint View Answer Discuss in Forum As per the given in question , we draw a figure of triangle ABC  
 ∠BPC = 120°
 ∴ ∠PBC + ∠PCB = 180°– 120°= 60°Correct Option: DAs per the given in question , we draw a figure of triangle ABC  
 ∠BPC = 120°
 ∴ ∠PBC + ∠PCB = 180°– 120°= 60°
 ∴ ∠ABC + ∠ACB = 2 × 60° = 120°
 ∴ ∠A = 180 – 120 = 60°
-  For a triangle ABC, D, E, F are the mid-points of its sides. If ∆ABC = 24 sq. units then ∆DEF is
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                        View Hint View Answer Discuss in Forum On the basis of given in question , we draw a figure triangle ABC ,  
 Here , ∆ABC = 24 sq. units
 From figure , we have∆DEF = 1 ∆ ABC 4 
 Correct Option: BOn the basis of given in question , we draw a figure triangle ABC ,  
 Here , ∆ABC = 24 sq. units
 From figure , we have∆DEF = 1 ∆ ABC 4 ∆DEF = 1 × 24 = 6 Sq. unit 4 
-  If I be the incentre of ∆ ABC and ∠B = 70° and ∠C = 50°, then the magnitude of ∠BIC is
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                        View Hint View Answer Discuss in Forum Firstly , We draw a figure of triangle ABC whose I is incentre ,  
 Here , ∠B = 70° and ∠C = 50°∠IBC = 70° = 35° ; 2 ∠ICB = 50° = 25° ; 2 
 Correct Option: CFirstly , We draw a figure of triangle ABC whose I is incentre ,  
 Here , ∠B = 70° and ∠C = 50°∠IBC = 70° = 35° ; 2 ∠ICB = 50° = 25° ; 2 
 We know that , ∠BIC + ∠ICB + ∠IBC = 180°
 ∴ ∠BIC = 180° – 35° – 25°
 ∠BIC = 180° – 60° = 120°
 
	