Let ABC and ABC' be two non-congruent triangles with sides AB = 4, AC = AC' = 2 \sqrt2 and angle B = 30 ^\circ . The absolute value of the difference between the areas of these triangles is \underline\hspace2cm .

Properties of Triangle PYQ — JEE Advanced 2009 · Shift 2 · Q37 with Solution

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Correct answer: 4

Understand the data

We have two non-congruent triangles $ABC$ and $ABC'$ such that:

$AB = 4$
$AC = AC' = 2\sqrt{2}$
$\angle B = 30^\circ$

This is the ambiguous case of triangle construction: with side $AB$ , side $AC$ , and angle $B$ fixed, there can be two different triangles.

Use the Law of Sines

In triangle $ABC$ , side $AC$ is opposite angle $B$ . So,

\frac{AC}{\sin B} = \frac{AB}{\sin C}

Substitute the values:

\frac{2\sqrt{2}}{\sin 30^\circ} = \frac{4}{\sin C}

Since $\sin 30^\circ = \tfrac12$ ,

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

4\sqrt{2} = \frac{4}{\sin C}

\sin C = \frac{1}{\sqrt{2}}

Hence,

C = 45^\circ \quad \text{or} \quad 135^\circ

These give the two non-congruent triangles.

Find the corresponding angle $A$ in each triangle

Using angle sum of a triangle:

A = 180^\circ - B - C

with $B = 30^\circ$ .

If $C = 45^\circ$ ,
$A = 180^\circ - 30^\circ - 45^\circ = 105^\circ$
If $C = 135^\circ$ ,
$A = 180^\circ - 30^\circ - 135^\circ = 15^\circ$

Area formula using sides $AB$ and $AC$ with included angle $A$

Area of a triangle:

\Delta = \frac12 \cdot AB \cdot AC \cdot \sin A

Here,

\Delta = \frac12 \cdot 4 \cdot 2\sqrt{2} \cdot \sin A = 4\sqrt{2}\sin A

So the two areas are:

For $A = 105^\circ$ : $\Delta_1 = 4\sqrt{2}\sin 105^\circ$
For $A = 15^\circ$ : $\Delta_2 = 4\sqrt{2}\sin 15^\circ$

Therefore,

|\Delta_1 - \Delta_2| = 4\sqrt{2}\,|\sin 105^\circ - \sin 15^\circ|

Simplify using identities

Since

\sin 105^\circ = \sin(90^\circ+15^\circ)=\cos 15^\circ,

we get

|\Delta_1 - \Delta_2| = 4\sqrt{2}(\cos 15^\circ - \sin 15^\circ)

Now use

\cos x - \sin x = \sqrt{2}\cos(x+45^\circ)

So for $x=15^\circ$ ,

\cos 15^\circ - \sin 15^\circ = \sqrt{2}\cos 60^\circ = \sqrt{2}\cdot \frac12 = \frac{1}{\sqrt{2}}

Hence,

|\Delta_1 - \Delta_2| = 4\sqrt{2}\cdot \frac{1}{\sqrt{2}} = 4

Final answer

The absolute difference between the areas is

\boxed{4}

More from Properties of Triangle

View written solutionFree

Correct answer: 4

Understand the data

We have two non-congruent triangles $ABC$ and $ABC'$ such that:

$AB = 4$
$AC = AC' = 2\sqrt{2}$
$\angle B = 30^\circ$

This is the ambiguous case of triangle construction: with side $AB$ , side $AC$ , and angle $B$ fixed, there can be two different triangles.

Use the Law of Sines

In triangle $ABC$ , side $AC$ is opposite angle $B$ . So,

\frac{AC}{\sin B} = \frac{AB}{\sin C}

Substitute the values:

\frac{2\sqrt{2}}{\sin 30^\circ} = \frac{4}{\sin C}

Since $\sin 30^\circ = \tfrac12$ ,

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

4\sqrt{2} = \frac{4}{\sin C}

\sin C = \frac{1}{\sqrt{2}}

Hence,

C = 45^\circ \quad \text{or} \quad 135^\circ

These give the two non-congruent triangles.

Find the corresponding angle $A$ in each triangle

Using angle sum of a triangle:

A = 180^\circ - B - C

with $B = 30^\circ$ .

If $C = 45^\circ$ ,
$A = 180^\circ - 30^\circ - 45^\circ = 105^\circ$
If $C = 135^\circ$ ,
$A = 180^\circ - 30^\circ - 135^\circ = 15^\circ$

Area formula using sides $AB$ and $AC$ with included angle $A$

Area of a triangle:

\Delta = \frac12 \cdot AB \cdot AC \cdot \sin A

Here,

\Delta = \frac12 \cdot 4 \cdot 2\sqrt{2} \cdot \sin A = 4\sqrt{2}\sin A

So the two areas are:

For $A = 105^\circ$ : $\Delta_1 = 4\sqrt{2}\sin 105^\circ$
For $A = 15^\circ$ : $\Delta_2 = 4\sqrt{2}\sin 15^\circ$

Therefore,

|\Delta_1 - \Delta_2| = 4\sqrt{2}\,|\sin 105^\circ - \sin 15^\circ|

Simplify using identities

Since

\sin 105^\circ = \sin(90^\circ+15^\circ)=\cos 15^\circ,

we get

|\Delta_1 - \Delta_2| = 4\sqrt{2}(\cos 15^\circ - \sin 15^\circ)

Now use

\cos x - \sin x = \sqrt{2}\cos(x+45^\circ)

So for $x=15^\circ$ ,

\cos 15^\circ - \sin 15^\circ = \sqrt{2}\cos 60^\circ = \sqrt{2}\cdot \frac12 = \frac{1}{\sqrt{2}}

Hence,

|\Delta_1 - \Delta_2| = 4\sqrt{2}\cdot \frac{1}{\sqrt{2}} = 4

Final answer

The absolute difference between the areas is

\boxed{4}

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Properties of Triangle question

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