Mathematics - Properties of Triangle Question with Solution | TestHub

The triangle ABC has $\angle A=60^{\circ },\angle B=45^{\circ }$. The bisector of $\angle A$ meets the side BC at T where $\mathrm{AT}=24$. Then which of the following is correct?

Options:

A. The length of BC is $18\sqrt{2}$.

B. The length of AC is less than AT.

C. The radius of the circumcircle is $12\sqrt{2}$.

D. The area of the triangle is $72(3+\sqrt{3})$ sq. units.

Correct Answer: C, D

Detailed explanation:

Given:

In $△ABC$, $\angle A=60^{\circ }$, $\angle B=45^{\circ }$.

AT is the angle bisector of $\angle A$, so $\angle BAT=\angle CAT=\frac{\angle A}{2}=\frac{60^{\circ }}{2}=30^{\circ }$.

AT = 24.

First, let's find $\angle C$:

In $△ABC$, $\angle A+\angle B+\angle C=180^{\circ }$

$60^{\circ }+45^{\circ }+\angle C=180^{\circ }$

$105^{\circ }+\angle C=180^{\circ }$

$\angle C=180^{\circ }-105^{\circ }=75^{\circ }$.

Now consider $△ABT$:

$\angle BAT=30^{\circ }$

$\angle B=45^{\circ }$

$\angle ATB=180^{\circ }-(\angle BAT+\angle B)=180^{\circ }-(30^{\circ }+45^{\circ })=180^{\circ }-75^{\circ }=105^{\circ }$.

Using the Sine Rule in $△ABT$:

$\frac{AT}{\sin B}=\frac{AB}{\sin \angle ATB}$

$\frac{24}{\sin 45^{\circ }}=\frac{AB}{\sin 105^{\circ }}$

$AB=\frac{24\sin 105^{\circ }}{\sin 45^{\circ }}$

We know $\sin 105^{\circ }=\sin (60^{\circ }+45^{\circ })=\sin 60^{\circ }\cos 45^{\circ }+\cos 60^{\circ }\sin 45^{\circ }$

$\sin 105^{\circ }=\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}+\frac{1}{2}\cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$.

And $\sin 45^{\circ }=\frac{\sqrt{2}}{2}$.

$AB=\frac{24\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)}{\frac{\sqrt{2}}{2}}=\frac{24(\sqrt{6}+\sqrt{2})}{4}\cdot \frac{2}{\sqrt{2}}=\frac{12(\sqrt{6}+\sqrt{2})}{\sqrt{2}}=12(\sqrt{3}+1)$.

Now consider $△ACT$:

$\angle CAT=30^{\circ }$

$\angle C=75^{\circ }$

$\angle ATC=180^{\circ }-(\angle CAT+\angle C)=180^{\circ }-(30^{\circ }+75^{\circ })=180^{\circ }-105^{\circ }=75^{\circ }$.

Since $\angle ATC=\angle C=75^{\circ }$, $△ACT$ is an isosceles triangle with $AC=AT$.

So, $AC=24$.

Let's evaluate the options:

Option A: The length of BC.

We need to find BT and TC.

From $△ABT$:

$\frac{BT}{\sin \angle BAT}=\frac{AT}{\sin B}$

$\frac{BT}{\sin 30^{\circ }}=\frac{24}{\sin 45^{\circ }}$

$BT=\frac{24\sin 30^{\circ }}{\sin 45^{\circ }}=\frac{24\cdot \frac{1}{2}}{\frac{\sqrt{2}}{2}}=\frac{12}{\frac{\sqrt{2}}{2}}=\frac{24}{\sqrt{2}}=12\sqrt{2}$.

From $△ACT$:

Since $AC=AT=24$ and $\angle ATC=\angle C=75^{\circ }$, we can find TC using the Sine Rule or by dropping a perpendicular.

Using the Sine Rule:

$\frac{TC}{\sin \angle CAT}=\frac{AT}{\sin \angle C}$

$\frac{TC}{\sin 30^{\circ }}=\frac{24}{\sin 75^{\circ }}$

$TC=\frac{24\sin 30^{\circ }}{\sin 75^{\circ }}=\frac{24\cdot \frac{1}{2}}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{12}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{48}{\sqrt{6}+\sqrt{2}}$

$TC=\frac{48(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}=\frac{48(\sqrt{6}-\sqrt{2})}{6-2}=\frac{48(\sqrt{6}-\sqrt{2})}{4}=12(\sqrt{6}-\sqrt{2})$.

$BC=BT+TC=12\sqrt{2}+12(\sqrt{6}-\sqrt{2})=12\sqrt{2}+12\sqrt{6}-12\sqrt{2}=12\sqrt{6}$.

So, option A, "The length of BC is $18\sqrt{2}$", is incorrect.

Option B: The length of AC is less than AT.

We found $AC=24$ and $AT=24$.

So, $AC=AT$.

Therefore, option B, "The length of AC is less than AT", is incorrect.

Option C: The radius of the circumcircle is $12\sqrt{2}$.

Let R be the circumradius of $△ABC$.

Using the Sine Rule in $△ABC$:

$\frac{AC}{\sin B}=2R$

$R=\frac{AC}{2\sin B}=\frac{24}{2\sin 45^{\circ }}=\frac{24}{2\cdot \frac{\sqrt{2}}{2}}=\frac{24}{\sqrt{2}}=12\sqrt{2}$.

So, option C, "The radius of the circumcircle is $12\sqrt{2}$", is correct.

Option D: The area of the triangle is $72(3+\sqrt{3})$ sq. units.

Area of $△ABC=\frac{1}{2}AB\cdot AC\sin A$

We have $AB=12(\sqrt{3}+1)$, $AC=24$, and $\sin A=\sin 60^{\circ }=\frac{\sqrt{3}}{2}$.

Area $=\frac{1}{2}\cdot 12(\sqrt{3}+1)\cdot 24\cdot \frac{\sqrt{3}}{2}$

Area $=6(\sqrt{3}+1)\cdot 12\sqrt{3}$

Area $=72\sqrt{3}(\sqrt{3}+1)$

Area $=72(3+\sqrt{3})$ sq. units.

So, option D, "The area of the triangle is $72(3+\sqrt{3})$ sq. units.", is correct.

The final answer is $\boxed{C,D}$.