Question

If one of the angles of a triangle is ${130}^{\circ }$ then the angle between the bisectors of the other two angles can be

(a) ${50}^{\circ }$

(b) ${65}^{\circ }$

(c) ${90}^{\circ }$

(d) ${155}^{\circ }$

Answer 1

ANSWER:
Let $△ABCbesuchthat\angle A={130}^{\circ }$
Here, BP is the bisector of $\angle BandCPisthebisectorof\angle C$.
∴ $\angle ABP=\angle PBC=1/2\angle B$
.....(1)
Also, $\angle ACP=\angle PCB=1/2\angle C$
.....(2)
In $△ABC$,
$\angle A+\angle B+\angle C={180}^{\circ }$
(Angle sum property)
⇒ ${130}^{\circ }+\angle B+\angle C={180}^{\circ }$
⇒ $\angle B+\angle C={180}^{\circ }-{130}^{\circ }={50}^{\circ }$
⇒ 1/2 $\angle B+1/2\angle C=1/2\times {50}^{\circ }={25}^{\circ }$
⇒ $\angle PBC+\angle PCB={25}^{\circ }$
.....(3)
[Using (1) and (2)]
In $△PBC$,
$\angle PBC+\angle PCB+\angle BPC={180}^{\circ }$
⇒${25}^{\circ }+\angle BPC={180}^{\circ }$
(Angle sum property)
[Using (3)]
⇒ $\angle BPC={180}^{\circ }-{25}^{\circ }={155}^{\circ }$
Thus, if one of the angles of a triangle is ${130}^{\circ }thentheanglebetweenthebisectorsoftheothertwoanglesis{155}^{\circ }$.
Hence, the correct answer is option (d).