Length of side of Triangle | PRMO II 2019 | Question 28

In a triangle ABC, it is known that \(\angle\)A=100\(^\circ\) and AB=AC. The internal angle bisector BD has length 20 units. Find the length of BC to the nearest integer, given that sin 10\(^\circ\)=0.174.

given, BD=20 units

\(\angle\)A=100\(^\circ\)

AB=AC

In \(\Delta\)ABD

\(\frac{BD}{sinA}=\frac{AD}{sin20^\circ}\)

or, \(\frac{BD}{sin100^\circ}=\frac{AD}{sin20^\circ}\)

or, 20=\(\frac{AD}{2sin10^\circ}\) or, AD=40sin10\(^\circ\)=6.96

In \(\Delta\)BDC

\(\frac{BD}{sin40^\circ}=\frac{BC}{sin120^\circ}=\frac{CD}{sin20^\circ}\)

or, CD=\(\frac{20}{2cos20^\circ}\)=\(\frac{20}{2 \times 0.9394}\)=10.65

So, AD+CD=AC=AB=17.6

since BD is angle bisector

\(\frac{BC}{AB}=\frac{CD}{AD}\)

or, BC=\(\frac{AB \times CD}{AD}\)=\(\frac{17.6 \times 10.65}{6.96}\)

=26.98=27.

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