Question : In $\triangle \mathrm{ABC}$, $\angle \mathrm{ABC} = 90^{\circ}$, $\mathrm{BP}$ is drawn perpendicular to $\mathrm{AC}$. If $\angle \mathrm{BAP} = 50^{\circ},$ what is the value of $\angle \mathrm{PBC}?$
Option 1: $30^{\circ}$
Option 2: $45^{\circ}$
Option 3: $50^{\circ}$
Option 4: $60^{\circ}$
Correct Answer: $50^{\circ}$
Solution :
In right-angled $\triangle \mathrm{ABC}$, where $\angle \mathrm{ABC} = 90^{\circ}$, $\mathrm{BP}$ is drawn perpendicular to $\mathrm{AC}$. Given that $\angle \mathrm{BAP }= 50^{\circ}$
In $\triangle\mathrm{ ABC}$,
$\mathrm{\angle C = 180^{\circ} - \angle A - \angle B = 180^{\circ} - 50^{\circ} - 90^{\circ} = 40^{\circ}}$
In $\mathrm{\triangle BPC}$,
$\mathrm{\angle PBC +\angle BPC +\angle PCB =180^{\circ}}$
⇒ $\mathrm{\angle PBC = 90^{\circ} - \angle PCB = 90^{\circ} - 40^{\circ} = 50^{\circ}}$
Hence, the correct answer is $50^{\circ}$.
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