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(B) In the right-angled triangle formed,the height of the building is $y$ (opposite side) and the distance from the base is $x$ (adjacent side).Using

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$x < y$

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(B) In the right-angled triangle formed,the height of the building is $y$ (opposite side) and the distance from the base is $x$ (adjacent side).Using the trigonometric ratio,$	an(70^\circ) = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{y}{x}$.We know that $	an(45^\circ) = 1$.Since the tangent function is increasing in the interval $(0^\circ, 90^\circ)$,and $70^\circ > 45^\circ$,it follows that $	an(70^\circ) > 	an(45^\circ)$.Therefore,$	an(70^\circ) > 1$.Substituting the ratio,$\frac{y}{x} > 1$.Since $x$ and $y$ are lengths (positive values),we can multiply both sides by $x$ to get $y > x$,which is equivalent to $x < y$.

$A$ person is watching from a point $A$ on the ground,which is $x \ m$ away from the base of a building. The height of the building is $y \ m$ and the angle of elevation of the top of the building is found to be $70^\circ$. Then:

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