Question : TF is a tower with point F on the ground. The angle of elevation of T from A is $\tan\;x^{\circ}=\frac{2}{5}$ and AF = 200 m. The angle of elevation of T from a nearer point B is $y^{\circ}$ with BF = 80 m. The value of $y$ is:
Option 1: 60$^{\circ}$
Option 2: 30$^{\circ}$
Option 3: 75$^{\circ}$
Option 4: 45$^{\circ}$
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Correct Answer: 45$^{\circ}$
Solution :
Let Height of the tower = $TF = h$ metre
$\angle TAF = x^{\circ}$, $\angle TBF = y^{\circ}$
$BF = 80$ metres and $AF = 200$ metres
In $\Delta AFT$,
$\tan x^{\circ} = \frac{TF}{AF}$
⇒ $\frac{2}5=\frac{h}{200}$
⇒ $ h=80$ m
in $\Delta BFT$,
$\tan y^{\circ} = \frac{TF}{FB}$
⇒ $\tan y^{\circ} =\frac{80}{80}$
⇒ $\tan y^{\circ}= 1$
So, $\tan y^{\circ} = \tan 45^{\circ}$
⇒ $y=45^{\circ}$
Hence, the correct answer is 45$^{\circ}$.
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