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A student sitting in a classroom sees a pictures on the black

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A student sitting in a classroom sees a pictures on the black board at a height of $1.5 m$ from the horizontal level of sight. The angle of elevation of the picture is $30^{o}$ . As the picture is not clear to him, he moves straight towards the black board and sees the picture at an angle of elevation $45^{o}$ . Find the distance moved by the student.

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A student sitting in a classroom sees a pictures on the black board at a height of

1.5 m

from the horizontal level of sight. The angle of elevation of the picture is

30^{o}

. As the picture is not clear to him, he moves straight towards the black board and sees the picture at an angle of elevation

45^{o}

. Find the distance moved by the student.

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Text solutionVerified

Let

A B

represent the board

D

represents I position of student

C

represents II position of student
In

△ A BC

\Rightarrow tan 45^{o} = \frac{A B}{A C}

\Rightarrow 1 = \frac{1.5}{y}

\Rightarrow y = 1.5

△ A BC

tan 30^{o} = \frac{A B}{A D}

\Rightarrow \frac{1}{3} = \frac{1.5}{x + y}

\Rightarrow x + y = 1.5 3

\Rightarrow x + 1.5 = 1.5 3

(

∵

y = 1.5

)

\Rightarrow x = 1.5 3 - 1.5 = 1.5 (3 - 1)

\Rightarrow x = 1.5 (1.732 - 1)

\Rightarrow x = 1.5 (0.732) = 1.098

m
The distance moved by the student

= 1.098

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Question Text	A student sitting in a classroom sees a pictures on the black board at a height of $1.5 m$ from the horizontal level of sight. The angle of elevation of the picture is $30^{o}$ . As the picture is not clear to him, he moves straight towards the black board and sees the picture at an angle of elevation $45^{o}$ . Find the distance moved by the student.
Answer Type	Text solution:1
Upvotes	150