Explanation
Then, ∠A+∠B+∠C+∠D=360∘.
To prove this, we join A and C, i.e. we draw the diagonal AC.
In △ABC,
∠CAB+∠ABC+∠BCA=180∘ [Sum of all angles of a triangle is 180∘].....(1).
In △ACD,
∠CAD+∠ADC+∠DCA=180∘ [Sum of all angle of a triangle is 180∘]....(2).
Adding (1) and (2), we get,
(∠CAB+∠ABC+∠BCA)+(∠CAD+∠ADC+∠DCA)=180∘+180∘
⟹ ∠ABC+∠ADC+(CAB+CAD)+(BCA+DCA)=360∘
⟹ ∠ABC+∠ADC+∠BAD+∠BCD=360∘
⟹ ∴∠A+∠B+∠C+∠D=360∘.
That is, the sum of all angles of a quadrilateral is 360o=4×90o, i.e. 4 right angles.
We know, by angle sum property, the sum of angles of a quadrilateral is 360o.
Given the ratio of angles of quadrilateral, ABCD is 3:4:5:6.
Let the angles of quadrilateral ABCD be 3x,4x,5x,6x, respectively.
We know, by angle sum property, the sum of angles of a quadrilateral is 360^\circ.
\Rightarrow 3x+4x+5x+6x=360^\circ
\Rightarrow 18x=360^\circ
\therefore x=20^\circ.
\angle A =3x
=3 \times 20^\circ
=60^\circ
\angle B =4x
=4\times 20^\circ
=80^\circ
\angle C =5x
=5 \times 20^\circ
=100^\circ
\angle D =6x
=6 \times 20^\circ
=120^\circ
The largest angle is equal to 120^\circ. Hence, option B is correct.
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