Triangles

Let the three angles be $2x, 3x,$ and $4x$.
By the Angle Sum Property of a Triangle:

$$2x + 3x + 4x = 180^\circ$$

$$9x = 180^\circ \implies x = \frac{180}{9} = 20^\circ$$

Calculate each angle:
• $\angle 1 = 2x = 2 \times 20^\circ = \mathbf{40^\circ}$
• $\angle 2 = 3x = 3 \times 20^\circ = \mathbf{60^\circ}$
• $\angle 3 = 4x = 4 \times 20^\circ = \mathbf{80^\circ}$
Since all three angles are strictly less than $90^\circ$, it is an Acute-angled Scalene Triangle.

Let the exterior angle be $\angle \text{ext} = 115^\circ$, and one interior opposite angle be $\angle 1 = 45^\circ$.
Step 1: Apply the Exterior Angle Theorem ($\angle \text{ext} = \angle 1 + \angle 2$):

$$115^\circ = 45^\circ + \angle 2$$

$$\mathbf{\angle 2 = 115^\circ - 45^\circ = 70^\circ}$$

Step 2: Find the third interior angle $\angle 3$ using the linear pair with the exterior angle:

$$\angle 3 = 180^\circ - 115^\circ = \mathbf{65^\circ}$$

(Or by Angle Sum: $180 - (45 + 70) = 180 - 115 = 65^\circ$).
The other two angles are $70^\circ$ and $65^\circ$.

• (a) $6\text{ cm}, 8\text{ cm}, 15\text{ cm}$:
Take the two smaller sides and add them: $6 + 8 = 14\text{ cm}$.
Compare with the third side: $14\text{ cm} < 15\text{ cm}$.
Since the sum of two sides is NOT greater than the third side ($6 + 8 \ngtr 15$), a triangle CANNOT be formed!
• (b) $5\text{ cm}, 12\text{ cm}, 13\text{ cm}$:
Check all pairs: $5 + 12 = 17 > 13$, $5 + 13 = 18 > 12$, and $12 + 13 = 25 > 5$.
Since the sum of any two sides is strictly greater than the third side, a triangle CAN be formed (in fact, it is a right-angled triangle since $5^2 + 12^2 = 13^2$).

The wall, ground, and ladder form a Right-Angled Triangle where:
Hypotenuse (ladder) $h = 15\text{ m}$
Base (distance from wall) $b = 9\text{ m}$
Perpendicular (height on wall) $p = ?$
By the Pythagoras Theorem ($h^2 = p^2 + b^2$):

$$15^2 = p^2 + 9^2$$

$$225 = p^2 + 81$$

$$p^2 = 225 - 81 = 144$$

$$p = \sqrt{144} = \mathbf{12\text{ meters}}$$

.
The ladder reaches a height of $12\text{ meters}$ up the wall.

Let the third side be $c$. By the Triangle Inequality Theorem:
1. The third side must be greater than the difference of the other two sides:

$$c > 11 - 8 \implies c > 3\text{ cm}$$

2. The third side must be less than the sum of the other two sides:

$$c < 11 + 8 \implies c < 19\text{ cm}$$

Therefore, the length of the third side must fall strictly between $3\text{ cm}$ and $19\text{ cm}$ ($3 < c < 19$).

North and East directions are mutually perpendicular ($90^\circ$).
The journey forms a right-angled triangle where:
Perpendicular $p = 12\text{ km}$
Base $b = 5\text{ km}$
Distance from start = Hypotenuse $h$
By the Pythagoras Theorem:

$$h^2 = p^2 + b^2 = 12^2 + 5^2 = 144 + 25 = 169$$

$$h = \sqrt{169} = \mathbf{13\text{ km}}$$

.
He is $13\text{ kilometers}$ from his starting point.

• Centroid ($G$): The point of concurrency where the three medians of a triangle intersect. It always lies strictly inside the triangle (for acute, right, and obtuse triangles alike) and divides each median in the ratio $2 : 1$.
• Orthocentre ($H$): The point of concurrency where the three altitudes of a triangle intersect.
• In an Obtuse-Angled Triangle: The orthocentre lies strictly OUTSIDE the triangle behind the obtuse-angled vertex, because two of the altitudes must be drawn to extended external bases.