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In a triangle,AD is a 2m high chalkboard which is 1 m above eye level BC,of a learner at C. BC=y.the angle of the top of the chalkboard,from C is B. ACD=aProve that y=2cosB.cos(B-a)/sina

In a triangle,AD is a 2m high chalkboard which is 1 m above eye level BC,of a learner at C. BC=y.the angle of the top of the chalkboard,from C is B. ACD=aProve that y=2cosB.cos(B-a)/sina

Grade:12

2 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
one year ago
Dear student

Kindly upload the question with proper notation or an image of the question so that it becomes clear to understand and solve

Regards
Sripad Sambrani
22 Points
one year ago
Given: AD=2m, AB=1m, BC=y, ∟ACB=b, ∟ACD=a
 
RTP: y=2*cosb*cos(b-a)/sina
 
Solution:
BD=(AD-AB)=2-1=1m, BD=AB=1m.
∟ACB=∟DCB, as AB=BD. hence ∟a=2∟b.
In triangle ABC, BC/AB=cotb
=> y/1=cotb
> y=cotb
 
RHS = 2*cosb*cos(b-a)/sina
= 2*cosb*cos(-b)/sin2b [since a=2b, b-2b=-b]
= 2*cosb*cosb/(2*sinb*cosb) [since cos(-θ)=cosθ]
= cotb = y = LHS Thus proved.
 
Trust this helps,
Regards,
SripadS

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