Advertisements
Advertisements
प्रश्न
BCDE is a square with side 90 cm and ∠F = 45°. The length of AF is ______.

पर्याय
\(60 \sqrt{2} \mathrm{~cm}\)
\(90 \sqrt{2} \mathrm{~cm}\)
\(120 \sqrt{2} \mathrm{~cm}\)
\(180 \sqrt{2} \mathrm{~cm}\)
Advertisements
उत्तर
BCDE is a square with side 90 cm and ∠F = 45°. The length of AF is \(\underline{\boldsymbol{180 \sqrt{2} \mathrm{~cm}}}\).
Explanation:
Since, BCDE is a square.
\[\therefore BE = CD = 90\ \text{cm}.\]
We know that,
\[\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}\]
From figure,
\[\angle AEB = \angle EFD = 45^\circ\] (Corresponding angles are equal)
In \[\triangle ABE,\]
\[\Rightarrow \tan 45^\circ = \frac{AB}{BE}\]
\[\Rightarrow 1 = \frac{AB}{90}\]
\[\Rightarrow AB = 90\ \text{cm}.\]
From figure,
\[AC = AB + BC = 90 + 90 = 180\ \text{cm}.\]
We know that,
\[\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}\]
In \[\triangle ACF,\]
\[\Rightarrow \sin 45^\circ = \frac{AC}{AF}\]
\[\Rightarrow \frac{1}{\sqrt{2}} = \frac{180}{AF}\]
\[\Rightarrow AF = 180\sqrt{2}\ \text{cm}.\]
