Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B

Q: Wait, why is $\sin C = \frac{\sqrt{3}}{2}$ leading to $120^\circ$ and not $60^\circ$?

That is a fantastic question! The equation $\sin C = \frac{\sqrt{3}}{2}$ has two solutions in the range $(0, 180^\circ)$: $60^\circ$ and $120^\circ$. However, the problem explicitly states that $\angle ACB$ is obtuse, meaning $C > 90^\circ$. Therefore, we must reject $60^\circ$ and accept $120^\circ$.

Q: Can I use Heron's formula instead of the Law of Cosines?

You absolutely could, but it would be much more difficult! Heron's formula requires the semi-perimeter $s$, which depends on the side $c$. Since you don't know $c$ yet, you would have to express $c$ in terms of the area, which essentially leads you back to the Law of Cosines or the Law of Sines. The Law of Cosines is the most direct path here.

Q: Why do we need the third side $c$ at all?

Great observation. The inradius $r$ is defined by the formula $r = \frac{\Delta}{s}$, where $\Delta$ is the area and $s$ is the semi-perimeter. Since $s = \frac{a+b+c}{2}$, you cannot calculate $s$ without knowing the length of the third side $c$. Finding $c$ is the essential bridge to the final answer.

Question

Consider a triangle $ABC$ and let $a, b$ and $c$ denote the lengths of the sides opposite to vertices $A, B$ and $C$ respectively. Suppose $a = 6, b = 10$ and the area of the triangle is $15\sqrt{3}$. If $\angle ACB$ is obtuse and if $r$ denotes the radius of the incircle of the triangle, then $r^2$ is equal to

Sigma Percentile · Accepted Answer

### The Geometric Odyssey: Unlocking the Incircle

Welcome, fellow traveler of the geometric realm. Today, we stand before a triangle $ABC$ that holds a secret.

We are given two sides, $a=6$ and $b=10$, and an area of $15\sqrt{3}$. We are also given the critical constraint that $\angle ACB$ is obtuse.

Our mission is to find $r^2$, where $r$ is the inradius. This is not just a calculation; it is a journey through the fundamental relationships that define a triangle.

### Phase 1: The Area Mystery

Every triangle is governed by its dimensions and the angles that bind them. We start with the most elegant of area formulas:

$$	ext{Area} = \frac{1}{2}ab \sin C$$

By substituting our known values, we get:

$$15\sqrt{3} = \frac{1}{2}(6)(10) \sin C$$

Simplifying this, we find $15\sqrt{3} = 30 \sin C$, which leads us to:

$$\sin C = \frac{\sqrt{3}}{2}$$

Now, pause. The sine function is a mirror; it gives the same value for an angle and its supplement. Thus, $\sin C = \frac{\sqrt{3}}{2}$ could mean $C = 60^\circ$ or $C = 120^\circ$.

The problem whispers a constraint: $C$ is obtuse. This is the filter that removes the ambiguity, forcing us to choose $C = 120^\circ$.

### Phase 2: The Missing Link

Now that we know $C = 120^\circ$, we have two sides and the included angle. To find the inradius, we need the perimeter, which requires the third side, $c$.

The Law of Cosines is our weapon of choice:

$$c^2 = a^2 + b^2 - 2ab \cos C$$

Let us substitute our values:

$$c^2 = 6^2 + 10^2 - 2(6)(10) \cos 120^\circ$$

Recall that $\cos 120^\circ = -\frac{1}{2}$. This negative sign is the heartbeat of the calculation, transforming the expression into:

$$c^2 = 36 + 100 - 120\left(-\frac{1}{2}ight)$$

$$c^2 = 136 + 60 = 196$$

Taking the square root, we find $c = 14$. The triangle is now fully revealed.

### Phase 3: The Incircle Connection

We are nearing the summit. The inradius $r$ is the radius of the circle that kisses all three sides of the triangle.

The relationship between the area $\Delta$, the semi-perimeter $s$, and the inradius $r$ is given by the beautiful formula:

$$r = \frac{\Delta}{s}$$

First, we find the semi-perimeter:

$$s = \frac{a+b+c}{2} = \frac{6+10+14}{2} = \frac{30}{2} = 15$$

Now, we simply divide the area by the semi-perimeter:

$$r = \frac{15\sqrt{3}}{15} = \sqrt{3}$$

The final step is to square this value as requested:

$$r^2 = (\sqrt{3})^2 = 3$$

And there it is—a perfect, elegant integer. We have navigated the trigonometry, conquered the Law of Cosines, and utilized the properties of the incircle to reach our destination. The final answer is **3**.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e T r ian g l e A B C$

$T h e A r e a F or m u l a$

$S u b s t i t u t in g t h e V a l u es$

$S o l v in g f or sin C$

$D e t er minin g t h e O b t u se A n g l e C$

$T h e C os in e R u l e f or S i d e c$

$S u b s t i t u t in g in t o t h e C os in e R u l e$

$C a l c u l a t in g S i d e c$

$F in d in g t h e S e mi - p er im e t er s$

$C a l c u l a t in g s$

$T h e I n r a d i u s F or m u l a$

$S u b s t i t u t in g f or r$

$C a l c u l a t in g t h e I n r a d i u s r$

$F ina l A n s w er : r^{2}$

The Sigma Insight: Properties of Triangles

The Geometric Odyssey

Phase 1

Phase 2

Phase 3

Similar Questions

Comprehension Passage

Let $a$ be the area of the triangle $A B C$ . Then the value of $(64 a)^{2}$ is

Then the inradius of the triangle $A B C$ is

If in a triangle $A B C$ , $A B = 5$ units, $∠ B = cos^{- 1} (\frac{3}{5})$ and radius of circumcircle of $Δ A B C$ is 5 units, then the area (in sq. units) of $Δ A B C$ is:

In $△ A B C$ , the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of $△ A B C$ is 30 cm $^{2}$ and R and r are respectively the radii of circumcircle and incircle of $△ A B C$ then the value of $2 R + r$ (in cm ) is equal to

Let $a, b$ and $c$ be the length of sides of a triangle $A B C$ such that $\frac{a + b}{7} = \frac{b + c}{8} = \frac{c + a}{9}$ . If $r$ and $R$ are the radius of incircle and radius of circumcircle of the triangle $A B C$ , respectively, then the value of $\frac{R}{r}$ is equal to

In a triangle PQR, let $∠ P QR = 3 0^{\circ}$ and the sides PQ and QR have lengths $103$ and 10, respectively. Then, which of the following statement(s) is (are) TRUE ?

In a triangle $A B C$ , let $∠ C = \frac{π}{2}$ . If $r$ is the inradius and $R$ is the circumradius of the triangle $A B C$ , then $2 (r + R)$ equals

In a triangle $X Y Z$ , let $x, y, z$ be the lengths of sides opposite to the angles $X, Y, Z$ , respectively, and $2 s = x + y + z$ . If $\frac{s - x}{4} = \frac{s - y}{3} = \frac{s - z}{2}$ and area of incircle of the triangle $X Y Z$ is $\frac{8 π}{3}$ , then

In a triangle $A B C$ , let $∠ C = π /2$ . If $r$ is the inradius and $R$ is the circumradius of the triangle, then $2 (r + R)$ is equal to

In a triangle the sum of two sides is $x$ and the product of the same sides is $y$ . If $x^{2} - c^{2} = y$ , where $c$ is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is

A circle is inscribed in an equilateral triangle of side $a$ . The area of any square inscribed in this circle is ..................

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Triangle ABC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Area Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting the Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for sinC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Determining the Obtuse Angle C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Cosine Rule for Side c

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting into the Cosine Rule

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Side c

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Semi−perimeter s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Inradius Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting for r

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Inradius r

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer: r2