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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

Equation

Algebra

Integers

## Check the Answer

But try the problem first…

Source

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

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