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Question

A triangle has sides 8 cm and 10 cm, and the angle between them is $120^\circ$. What is its area?

This question was previously asked in
SSC CGL 2025 Tier 1 Question Paper (25-Sep-2025) (Shift 3)
The correct answer is
$20\sqrt{3}\text{ cm}^2$

Calculating Triangle Area Using Sides and Angle

The area of a triangle can be calculated when two sides and the angle between them are known. The formula used is:

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

Where 'a' and 'b' are the lengths of the two sides, and 'C' is the measure of the included angle.

Applying the Formula

  • Given sides: $a = 8$ cm, $b = 10$ cm
  • Given included angle: $C = 120^\circ$
  • Substitute these values into the formula:

Area $= \frac{1}{2} \times 8 \text{ cm} \times 10 \text{ cm} \times \sin(120^\circ)$

Calculating the Sine Value

The sine of $120^\circ$ is a standard trigonometric value:

$\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$

Final Area Calculation

Now, substitute the sine value back into the area calculation:

Area $= \frac{1}{2} \times 8 \times 10 \times \frac{\sqrt{3}}{2}$

Area $= 4 \times 10 \times \frac{\sqrt{3}}{2}$

Area $= 40 \times \frac{\sqrt{3}}{2}$

Area $= 20\sqrt{3} \text{ cm}^2$

The area of the triangle is $20\sqrt{3} \text{ cm}^2$.

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