In a triangle ABC, andang;B = 90anddeg;, D is a point on line AB such that it divides AB into two equal parts. AB is denoted by 2x and AC by y, then find the value of sin andtheta; where andang;DCB = andtheta;anddeg;.a)2x/(y2andndash; 3x2)b)x/(y2andndash; 3x2)c)2x/andradic;(y2andndash; 3x2)d)x/andradic;(y2andndash; 3x2)Correct answer is option 'D'. Can you explain this answer?

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In a triangle ABC, ∠B = 90°, D is a point on line AB such that it divides AB into two equal parts. AB is denoted by 2x and AC by y, then find the value of sin θ where ∠DCB = θ°.

a)
2x/(y² – 3x²)
b)
x/(y² – 3x²)
c)
2x/√(y² – 3x²)
d)
x/√(y² – 3x²)

Correct answer is option 'D'. Can you explain this answer?

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In a triangle ABC, ∠B = 90°, D is a point on line AB such that...

Appling Pythagoras theorem on triangle ABC
AC² = AB² + BC²
⇒ y² = (2x)² + BC²
⇒ BC² = y² – 4x²
Now applying Pythagoras theorem on triangle DBC
DC² = DB² + BC²
⇒ z² = x² + BC²
⇒ BC² = z² – x²
Now, on comparing the value of BC² we get
y² – 4x² = z² – x²
∴ z² = y² – 3x²

∴ z = √(y² – 3x²)
Now in Triangle DBC,
sin θ = DB/DC
∴ sin θ = x/z
∴ sin θ = x/√(y² – 3x²)

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In a triangle ABC, ∠B = 90°, D is a point on line AB such that it divides AB into two equal parts. AB is denoted by 2x and AC by y, then find the value of sin θ where ∠DCB = θ°.a)2x/(y2– 3x2)b)x/(y2– 3x2)c)2x/√(y2– 3x2)d)x/√(y2– 3x2)Correct answer is option 'D'. Can you explain this answer?

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