Question

as shwon in figure,LK=6√2 then 1)Mk=? 2)ML=? 3)MN=?​

Answer 1

Given :-

• LK = 6√2 .

To Find :-

• MK = ?
• ML = ?
• MN = ?

Solution :-

In right angled ∆MLK we have,

→ cos θ = LK / MK

→ cos 30° = 6√2/ MK

→ (√3/2) = 6√2/MK

→ √3 * MK = 12√2

→ MK = (12√2/√3)

→ MK = (4 * 3 * √2)/√3

→ MK = 4 * √3 * √2

→ MK = 4√6 (Ans.)

also, In right angled ∆MLK we have,

→ tan θ = LM / LK

→ (1/√3) = LM / 6√2

→ LM = 6√2/√3

→ LM = (6√2/√3) * (√3/√3)

→ LM = (6√6/3)

→ LM = 2√6 (Ans.)

now, in right angled ∆MKN,

→ sin θ = MK / MN

→ sin 45° = 4√6/ MN

→ (1/√2) = 4√6 / MN

→ MN = 4 * √3 * √2 * √2

→ MN = 4 * 2 * √3

→ MN = 8√3 (Ans.)

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