Question

If ABCD is a quadrilateral and AP and DP are bisectors of \[\angle A\] and \[\angle D\]. The value of \[x\] is:

A.\[60^\circ \]
B.\[85^\circ \]
C.\[95^\circ \]
D.\[100^\circ \]

Answer 1

Hint: Here, we will find the value of the required angle inside the quadrilateral using the properties of the quadrilateral and will also use the properties of the triangle. We will use the fact that the sum of all internal angles of a quadrilateral is equal to \[360^\circ \] and the sum of angles of a triangle is equal to \[180^\circ \].

Complete step by step solution:
It is given that the \[AP\] bisects \[\angle A\] and \[DP\] bisects \[\angle D\].
Therefore,
\[\begin{array}{l}\angle DAP = \angle BAP\\\angle ADP = \angle PDC\end{array}\]
We know, \[\angle DAP + \angle BAP = \angle A\]
Substituting \[\angle DAP = \angle BAP\] in the above equation, we get
\[ \Rightarrow \angle DAP + \angle DAP = \angle A\]
Adding the like terms, we get
\[ \Rightarrow 2\angle DAP = \angle A\]
Dividing both sides by 2, we get
\[ \Rightarrow \angle DAP = \dfrac{1}{2}\angle A\] ……….. \[\left( 1 \right)\]
Similarly,
\[ \Rightarrow \angle ADP = \dfrac{1}{2}\angle B\] …….. \[\left( 2 \right)\]
We know that the sum of all internal angles of a quadrilateral is equal to \[360^\circ \]. We will add all the angles of quadrilateral \[ABCD\] and we will equate it with \[360^\circ \].
\[ \Rightarrow \angle A + \angle B + \angle C + \angle D = 360^\circ \]
Substituting the value of all the angles, we get
\[ \Rightarrow \angle A + 130^\circ + 60^\circ + \angle D = 360^\circ \]
Using equation \[\left( 1 \right)\] and \[\left( 2 \right)\], we get
\[ \Rightarrow 2\angle DAP + 130^\circ + 60^\circ + 2\angle ADP = 360^\circ \]
On further simplification, we get
\[ \Rightarrow 2\angle DAP + 2\angle ADP + 190^\circ = 360^\circ \]
Subtracting \[190^\circ \] from both sides, we get
\[ \Rightarrow 2\angle DAP + 2\angle ADP + 190^\circ - 190^\circ = 360^\circ - 190^\circ \]
On simplifying the terms, we get
\[ \Rightarrow \angle DAP + \angle ADP = 85^\circ \] …….. \[\left( 3 \right)\]
We know that the sum of angles of a triangle is equal to \[180^\circ \].
Therefore, in \[\Delta APD\]
\[\angle DAP + \angle ADP + \angle APD = 180^\circ \]
Using equation \[\left( 3 \right)\], we can rewrite above equation as
\[ \Rightarrow 85^\circ + x = 180^\circ \]
Subtracting \[85^\circ \] from sides, we get
\[\begin{array}{l} \Rightarrow 85^\circ + x - 85^\circ = 180^\circ - 85^\circ \\ \Rightarrow x = 95^\circ \end{array}\]
Hence, the value of \[x\] is \[95^\circ \].
Thus, the correct option is option C.

Note: Here we have obtained the angle inside the quadrilateral. The sum of all interior angles of a quadrilateral is four times the right angle. The diagonals of a quadrilateral intersect each other. The sum of all exterior angles of a quadrilateral is also four times the right angle.