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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

Divhithani , 7 Years ago
Grade 12
anser 2 Answers
Saurabh Koranglekar

Last Activity: 5 Years 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

Last Activity: 5 Years 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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