Let \(ABC\) and \(ABC^{\prime}\) be two non-congruent triangles with sides \(AB=4\), \(AC=AC^{\prime}=2 \sqrt{2}\) and angle \(B=30^{\circ}\). The absolute value of the difference between the areas of these triangles is

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Properties of Triangle

Let

A B C

and

A B C^{'}

be two non-congruent triangles with sides

A B = 4

A C = A C^{'} = 22

and angle

B = 3 0^{\circ}

. The absolute value of the difference between the areas of these triangles is

Answer: 4

Using the sine rule in triangle

A B C

\frac{A C}{s i n B} = \frac{A B}{s i n C}

\frac{2 2}{s i n 3 0 ^{\circ}} = \frac{4}{s i n C}

, so

sin C = \frac{4 \cdot 1/2}{2 2} = \frac{2}{2}

.

Thus

C = 4 5^{\circ}

13 5^{\circ}

, giving two non-congruent triangles.

For

C = 4 5^{\circ}

A = 18 0^{\circ} - 3 0^{\circ} - 4 5^{\circ} = 10 5^{\circ}

.

For

C = 13 5^{\circ}

A = 18 0^{\circ} - 3 0^{\circ} - 13 5^{\circ} = 1 5^{\circ}

.

Using the sine rule to find

B C

\frac{B C}{sin A} = \frac{A B}{sin C} = \frac{4}{2 /2} = 42

For the first triangle:

B C_{1} = 42 sin 10 5^{\circ}

.

For the second triangle:

B C_{2} = 42 sin 1 5^{\circ}

.

The area of each triangle is

\frac{1}{2} \cdot A B \cdot B C \cdot sin B = \frac{1}{2} \cdot 4 \cdot B C \cdot \frac{1}{2} = B C

.

The difference in areas is

∣ B C_{1} - B C_{2} ∣ = 42 ∣ sin 10 5^{\circ} - sin 1 5^{\circ} ∣ = 42 \cdot \frac{2}{2} = 4

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Integer answer