Question

ami didi standing on a railway Over Bridge of 5√3 m. height observed the engine of the train from one side of the bridge at an angel of depression of 30° .But just after 2 second she observed the engine at an angel of depression pf 45° from the other side of the bridge.find the speed of the train in metre per second

Answer 1

❏ Some important FormulaS:-

For a right angled triangle∆ABC,

if $\angle B=90\degree$ and AC is hypotenuse,

And if $\angle C=\theta$,

$\sf\longrightarrow\sin \theta=\frac{AB}{AC}$

$\sf\longrightarrow\cos \theta=\frac{BC}{AC}$

$\sf\longrightarrow\tan \theta=\frac{AB}{BC}$

$\sf\longrightarrow\cosec \theta=\frac{AC}{AB}$

$\sf\longrightarrow\sec \theta=\frac{AC}{BC}$

$\sf\longrightarrow\cot \theta=\frac{BC}{AB}$

AND,

$\sf\longrightarrow velocity=\frac{distance}{time}$

❏ Question:-

Q) Ami didi standing on a railway Over Bridge of 5√3 m. height observed the engine of the train from one side of the bridge at an angel of depression of 30° .But just after 2 second she observed the engine at an angel of depression 45° from the other side of the bridge.find the speed of the train in metre per second.

❏ Solution:-

(•)Given➾

•Height(h) of the Bridge=AB=$5\sqrt{3}\:m$

• Angle of depression from one side= 30°=$\angle ACB$

• Angle of depression from other side= 45°=$\angle ADB$

• required time (t)=2 sec.

(•)To find:-

➾ Speed of the train=?

(•)solution:-

Now, from the ∆ADB,

$\sf\longrightarrow\tan 45\degree=\frac{AB}{BD}$

$\sf\longrightarrow 1=\frac{5\sqrt{3}}{BD}$

$\sf\longrightarrow\bf BD=5\sqrt{3}\:m$

Now, from the ∆ACB,

$\sf\longrightarrow\tan 30\degree=\frac{AB}{BC}$

$\sf\longrightarrow \frac{1}{\sqrt{3}}=\frac{5\sqrt{3}}{BC}$

$\sf\longrightarrow\bf BC=5/times\sqrt{3}\times\sqrt{3}\:m$

$\sf\longrightarrow\bf BC=5/times3\:m$

$\sf\longrightarrow\bf\bf BC=15\:m$

$\therefore CD=BC+BD$

$\therefore CD=(15+5\sqrt{3})\:m$

Therefore, distance covered by the train in 2 seconds is =(15+5√3) m

$\sf\therefore speed \:of \:the \:train=\frac{15+5\sqrt{3}}{2}$

$\sf\longrightarrow \boxed{\red{speed \:of \:the \:train}=11.83\:m.s{}^{-1}}\:\:(almost)$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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