Question

in triangle ABC and XYZ if angle A and X are acute angles such that cos A=cos X then show that angle A=X​

Answer 1

Answer:

Given: cos A = cos X

cos A = Adjacent/Hypotenuse = AB/AC,  cos X = Adjacent/Hypotenuse = XY/XZ

∴ AB/AC = XY/XZ

Let: AB/AC = XY/XZ = k --- Eq (1)

∴ By Pythagoras Theorem,

= BC/YZ =  --- Eq (2)

=  From Eq (1),

= AB = kAC --- Eq (3)

=  = kYZ --- Eq (4)

∴ Put Eq (3) and Eq (4) in Eq (2)

= BC/YZ =  $\sqrt{AC^2-k^2A}/\sqrt{XZ^2-k^2XZ}$

∴ BC/YZ = $\sqrt{AC^2(1-k^2)} /\sqrt{XZ^2(1-k^2)}$[∵ √AB = √A x √B]

= √AC² x √1 - k²/√XZ² x √1 - k² (√,² and √1 - k² gets Cancelled)

∴ BC/YZ = AC/XZ = AB/XY --- Eq (5)

From Eq (5), ΔABC = ΔXYZ

= ∠A = ∠X

Hence Proved