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Byju's Answer

Standard XII

Mathematics

Angle between Two Lines

Let ABCD be a...

Question

Let $A B C D$ be a square. $E$ and $F$ be points on $A C$ such that $A E = E F = F C = \frac{A C}{3}$ . Then $tan (D E B F)$ equals:

\frac{3}{4}

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\frac{1}{\sqrt{3}}

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\frac{1}{2}

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\frac{1}{3}

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Solution

The correct option is A $\frac{3}{4}$

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Similar questions

Let ABCD be a square. E and F be points on AC such that AE = EF = FC =AC/3. Then tan (ÐEBF) equals :

Q. In trapezium

A B C D

A B | | D C

E

and

F

are points on non-parallel sides

A D

and

B C

respectively such that

E F ∥ A B

. Prove that

\frac{A E}{E D} = \frac{B F}{F C}

ABCD is a trapezium with $A B ∥ D C$ . E and F are points on non-parallel sides AD and BC respectively such that $E F ∥ A B$ . Show that $\frac{A E}{E D} = \frac{B F}{F C}$ .

Q. ABCD is trapezium with AB||DC. E and F are points on non-parralel sides AD and BC respectively such that EF||AB. Show that AE/ED=BF/FC.

State true or false:

The alongside figure shows a parallelogram

A B C D

in which

A E = E F = F C

, then

D E B F

is a parallelogram.

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Let ABCD be a square. E and F be points on AC such that AE=EF=FC=AC3. Then tan(DEBF) equals:

The correct option is A 34

Let $A B C D$ be a square. $E$ and $F$ be points on $A C$ such that $A E = E F = F C = \frac{A C}{3}$ . Then $tan (D E B F)$ equals:

The correct option is A $\frac{3}{4}$