Question

In a triangle ABC with $\angle A=90$, P is a point on BC such that $PA:PB=3:4$. If $AB=\sqrt{7}$ and $AC=\sqrt{5}$ , then BP : PC is

• A. $2:1$
• B. $4:3$
• C. $4:5$
• D. $8:7$

Answer 1

The correct option is A $2:1$

Let $PA=3x$, and $PB=4x$

$BC=\sqrt{{\sqrt{7}}^2+{\sqrt{5}}^2}=\sqrt{12}PC=BC-PB=\sqrt{12}-4x$

Using cosine rule for $△ABC$

$\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{7+12-5}{2\sqrt{12\times 7}}$

$=\frac{\sqrt{7}}{\sqrt{12}}\dots \dots \left(1\right)$

Similarly, $△APB$

$\cos B=\frac{{AB}^2+{BP}^2-{AP}^2}{2AB\times BP}$

$\cos B=\frac{7+16x^2-9x^2}{2\sqrt{7}\times 4x}=\frac{7(1+x^2)}{8x\sqrt{7}}\dots \dots \left(2\right)$

$\left(1\right)=\left(2\right)$

$\sqrt{\frac{7}{12}}=\frac{\sqrt{7}(1+x^2)}{8x}$

$\Rightarrow 4x=\sqrt{3}(1+x^2)\Rightarrow (x-\sqrt{3})(\sqrt{3}x-1)=0\Rightarrow x=\sqrt{3},\frac{1}{\sqrt{3}}$

If $x=\sqrt{3}⟹4x>\sqrt{12}$ which is not possible

$\therefore x=\frac{1}{\sqrt{3}}$

$\therefore \frac{BP}{PC}=\frac{4x}{\sqrt{12}-4x}$

$=\frac{4/\sqrt{3}}{2\sqrt{3}-4/\sqrt{3}}=\frac{2}{1}$