In a triangle ABC- if AB-2x-1- AC- 2 -2x and BC-2x-1- Find the angle - BAC-

AB^2 = (2x+1)^2 = 4x^2+1 4x AC^2 = ( 2√(2x) nbsp;)^2 = 8x BC^2 = (2x+1)^2 = 4x^2+1+4x We can observe here, AB^2+AC^2=4x^2+1 4x+8x = 4x^2+1+4x =BC^2 AB ...

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Standard X

Mathematics

Converse of Pythagoras' Theorem

In a triangle...

Question

In a triangle ABC, if AB=2x-1, $A C = 2 \sqrt{2 x}$ and BC=2x+1. Find the angle $∠ B A C$ .

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Solution

$A B^{2} = (2 x + 1)^{2} = 4 x^{2} + 1 - 4 x A C^{2} = (2 \sqrt{2 x})^{2} = 8 x B C^{2} = (2 x + 1)^{2} = 4 x^{2} + 1 + 4 x$
We can observe here,
$A B^{2} + A C^{2} = 4 x^{2} + 1 - 4 x + 8 x = 4 x^{2} + 1 + 4 x = B C^{2}$
$A B^{2} + A C^{2} = B C^{2}$
So from converse of Pythagorean theorem we have,
$∠ B A C = 90^{\circ}$

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In a triangle ABC, if AB=2x-1, AC=2√2x and BC=2x+1. Find the angle ∠BAC.

AB2=(2x+1)2=4x2+1−4xAC2=(2√2x)2=8xBC2=(2x+1)2=4x2+1+4x We can observe here, AB2+AC2=4x2+1−4x+8x=4x2+1+4x=BC2 AB2+AC2=BC2 So from converse of Pythagorean theorem we have, ∠BAC=90∘

In a triangle ABC, if AB=2x-1, $A C = 2 \sqrt{2 x}$ and BC=2x+1. Find the angle $∠ B A C$ .