A) \[A\to (i),\,B\to (iv);\,C\to (ii);\,D\to (iii)\]
B) \[A\to (iv),\,B\to (ii);\,C\to (i);\,D\to (iii)\]
C) \[A\to (iv),\,B\to (i);\,C\to (ii);\,D\to (iii)\]
D) \[A\to (i),\,B\to (iv);\,C\to (iii);\,D\to (ii)\]
Correct Answer: B
Solution :
A. \[\because \] sum of the angles in a triangle \[=\text{ }180{}^\circ \] i.e., \[{{65}^{\text{o}}}+{{85}^{\text{o}}}+x={{180}^{\text{o}}}\Rightarrow \,x={{30}^{\text{o}}}\] B. Let the angles be 3x, 4x, 5x and 6x. We have \[3x\text{ }+\text{ }4x\text{ }+\text{ }5x\text{ }+\text{ }6x\text{ }=\text{ }360{}^\circ \] \[18x={{360}^{\text{o}}},\,\,x={{20}^{\text{o}}}\] Difference between greatest and smallest angles \[=\text{ }6x-3x=\text{ }3x=\text{ }3\times 20{}^\circ \text{ }=60{}^\circ \]. C. We have, x + x ? 10 + x + 30 + 2x = \[360{}^\circ \] = 5x + 20 = 360 \[x=\frac{340}{5}\,={{68}^{\text{o}}}\] The angles are \[68{}^\circ \], \[68{}^\circ \]-\[10{}^\circ \], \[68{}^\circ +30{}^\circ \], \[2\times 68{}^\circ \], i.e.\[68{}^\circ \], \[58{}^\circ \], \[98{}^\circ \], \[136{}^\circ \] \[\therefore \] Greatest angle =\[136{}^\circ \]. D. In a parallelogram, opposite angles are equal. So \[\angle A=\angle C\Rightarrow \,\angle A-\angle C={{0}^{\text{o}}}\]
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