Try this beautiful problem from the Pre-RMO II, 2019, Question 28, based on Length of side of triangle.

## Length of side of triangle – Problem 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.

- is 107
- is 27
- is 840
- cannot be determined from the given information

**Key Concepts**

Equation

Algebra

Integers

## Check the Answer

But try the problem first…

Answer: is 27.

PRMO II, 2019, Question 28

Higher Algebra by Hall and Knight

## Try with Hints

First hint

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

Second Hint

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

Final Step

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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

Google