In a triangle ABC with angle A = 90^{circ},P | vedclass

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Question

(A) Given,$	riangle ABC$ is a right-angled triangle with $\angle A = 90^{\circ}$.$AC = \sqrt{5}$,$AB = \sqrt{7}$.In $	riangle ABC$,$BC^2 = AB^2 + AC

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

(A) Given,$	riangle ABC$ is a right-angled triangle with $\angle A = 90^{\circ}$.$AC = \sqrt{5}$,$AB = \sqrt{7}$.In $	riangle ABC$,$BC^2 = AB^2 + AC^2 = 7 + 5 = 12$,so $BC = \sqrt{12} = 2\sqrt{3}$.Let $PA = 3x$ and $PB = 4x$. In $	riangle ABP$,by the Law of Cosines:$PA^2 = AB^2 + PB^2 - 2(AB)(PB)\cos B$$(3x)^2 = (\sqrt{7})^2 + (4x)^2 - 2(\sqrt{7})(4x)\cos B$$9x^2 = 7 + 16x^2 - 8x\sqrt{7}\cos B$Since $\cos B = \frac{AB}{BC} = \frac{\sqrt{7}}{2\sqrt{3}}$,we have:$9x^2 = 7 + 16x^2 - 8x\sqrt{7} \left(\frac{\sqrt{7}}{2\sqrt{3}}ight)$$9x^2 = 7 + 16x^2 - \frac{28x}{\sqrt{3}}$$7x^2 - \frac{28x}{\sqrt{3}} + 7 = 0$$x^2 - \frac{4x}{\sqrt{3}} + 1 = 0$$\sqrt{3}x^2 - 4x + \sqrt{3} = 0$$\sqrt{3}x^2 - 3x - x + \sqrt{3} = 0$$\sqrt{3}x(x - \sqrt{3}) - 1(x - \sqrt{3}) = 0$$(x - \sqrt{3})(\sqrt{3}x - 1) = 0$Since $P$ lies on $BC$,$PB $PB = 4x = \frac{4}{\sqrt{3}}$.$PC = BC - PB = 2\sqrt{3} - \frac{4}{\sqrt{3}} = \frac{6-4}{\sqrt{3}} = \frac{2}{\sqrt{3}}$.Therefore,$BP:PC = \frac{4}{\sqrt{3}} : \frac{2}{\sqrt{3}} = 2:1$.

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

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