Q195 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
In a triangle $ABC$, two sides $BC$ and $CA$ are in the ratio $2:1$ and their opposite corresponding angles are in the ratio $3:1$.
One of the angles of the triangle is
In a triangle $ABC$, two sides $BC$ and $CA$ are in the ratio $2:1$ and their opposite corresponding angles are in the ratio $3:1$.
One of the angles of the triangle is
- A$15°$
- Correct answerB$30°$
- C$45°$
- D$75°$
Solution
Let $BC = a$ and $CA = b$. Given $a:b = 2:1$, so $a = 2b$.
Let the angles opposite to $BC$ and $CA$ be $A$ and $B$ respectively. Given $A:B = 3:1$, so $A = 3B$.
Using the sine rule in $\Delta ABC$:
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$
$\dfrac{2b}{\sin 3B} = \dfrac{b}{\sin B}$
$2 \sin B = \sin 3B$
$2 \sin B = 3 \sin B - 4 \sin^3 B$
$4 \sin^3 B - \sin B = 0$
$\sin B (4 \sin^2 B - 1) = 0$
Since $B$ is an angle of a triangle, $\sin B \neq 0$.
$4 \sin^2 B = 1 \Rightarrow \sin B = \dfrac{1}{2}$
$B = 30^\circ$ (as $B = 150^\circ$ would make $A = 450^\circ$, which is not possible).
The angles of the triangle are $A = 90^\circ$, $B = 30^\circ$, and $C = 60^\circ$.
One of the angles is $30^\circ$.
Answer: $30°$
Let the angles opposite to $BC$ and $CA$ be $A$ and $B$ respectively. Given $A:B = 3:1$, so $A = 3B$.
Using the sine rule in $\Delta ABC$:
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$
$\dfrac{2b}{\sin 3B} = \dfrac{b}{\sin B}$
$2 \sin B = \sin 3B$
$2 \sin B = 3 \sin B - 4 \sin^3 B$
$4 \sin^3 B - \sin B = 0$
$\sin B (4 \sin^2 B - 1) = 0$
Since $B$ is an angle of a triangle, $\sin B \neq 0$.
$4 \sin^2 B = 1 \Rightarrow \sin B = \dfrac{1}{2}$
$B = 30^\circ$ (as $B = 150^\circ$ would make $A = 450^\circ$, which is not possible).
The angles of the triangle are $A = 90^\circ$, $B = 30^\circ$, and $C = 60^\circ$.
One of the angles is $30^\circ$.
Answer: $30°$
Hence, the correct answer is option: 'B'.