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UPSC CDS 2 2022 Math Paper

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Question : 79 of 100

Marks: +1, -0

In a triangle ABC,

\text{angle } A = 60^{\circ}

AB^{2} + AC^{2} - BC^{2}

equal to ?

AB.AC
AB.BC
AC.BC
2AB.AC

Solution:

Concept:

Use the cosine rule to relate the sides of the triangle with the given angle.

Explanation:

In

\text{triangle } ABC

,

\text{angle } A = 60^{\circ}

.

By the cosine rule:

\cos A = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC}

.

Substitute

\cos 60^{\circ} = \frac{1}{2}

.

Then

\frac{1}{2} = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC}

.

Cross-multiply:

AB^2 + AC^2 - BC^2 = AB \cdot AC

.

Answer:

AB \cdot AC

(Option A).

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