Menu Close

the-opposite-side-of-a-triangle-is-x-y-and-the-hypotenuse-is-3x-y-Given-that-that-A-is-an-acute-angle-find-the-value-of-Cos-A-




Question Number 36239 by Rio Mike last updated on 30/May/18
the opposite side of a triangle is    x + y,and the hypotenuse is 3x+y  Given that that A is an acute angle  find the value of Cos A
$$\mathrm{the}\:\mathrm{opposite}\:\mathrm{side}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{is}\: \\ $$$$\:{x}\:+\:{y},\mathrm{and}\:\mathrm{the}\:\mathrm{hypotenuse}\:\mathrm{is}\:\mathrm{3}{x}+{y} \\ $$$$\mathrm{Given}\:\mathrm{that}\:\mathrm{that}\:\mathrm{A}\:\mathrm{is}\:\mathrm{an}\:\mathrm{acute}\:\mathrm{angle} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{Cos}\:\mathrm{A} \\ $$
Commented by Rasheed.Sindhi last updated on 30/May/18
 Opposite side to ∠A=x+y  Hypotenuse=3x+y  Adjacent side to ∠A=(√((3x+y)^2 −(x+y)^2 ))               =(√(9x^2 +6xy+y^2 −x^2 −2xy−y^2 ))               =(√(8x^2 +4xy))               =2(√(x(2x+y)))  cos A=((2(√(x(2x+y))))/(3x+y))
$$\:\mathrm{Opposite}\:\mathrm{side}\:\mathrm{to}\:\angle\mathrm{A}=\mathrm{x}+\mathrm{y} \\ $$$$\mathrm{Hypotenuse}=\mathrm{3x}+\mathrm{y} \\ $$$$\mathrm{Adjacent}\:\mathrm{side}\:\mathrm{to}\:\angle\mathrm{A}=\sqrt{\left(\mathrm{3x}+\mathrm{y}\right)^{\mathrm{2}} −\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{9x}^{\mathrm{2}} +\mathrm{6xy}+\mathrm{y}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} −\mathrm{2xy}−\mathrm{y}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{8x}^{\mathrm{2}} +\mathrm{4xy}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\sqrt{\mathrm{x}\left(\mathrm{2x}+\mathrm{y}\right)} \\ $$$$\mathrm{cos}\:\mathrm{A}=\frac{\mathrm{2}\sqrt{\mathrm{x}\left(\mathrm{2x}+\mathrm{y}\right)}}{\mathrm{3x}+\mathrm{y}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *