tan Θ =  AB 
BQ 
tan Θ =  h  ...(i) 
b 
tan (90° – Θ) =  AB 
BP 
cot Θ =  h  ...(ii) [∵ tan (90° – Θ) = cot Θ] 
a 
tan Θ cot Θ =  h  ×  h 
b  a 
1 =  h^{2}  [∵ tan Θ cot Θ = 1] 
ab 
h =  ab 
tan 30° =  AB 
BC 
1  =  h 
3  100 
h =  100  m 
3 
tan Θ =  AB  =  h  ...(i) 
BC  100 
tan 2Θ =  AB 
BD 
2tanΘ  =  h  [  ∵ tan 2Θ =  2tanΘ  ] 
1 – tan^{2} Θ  60  1 – tan^{2} Θ 
120 ×  h  =  h  [  1 –  (  h  )  ^{2}  ]  [From eq. (i)] 
160  160 
3  = 1 –  (  h  )  ^{2} 
4  160 
(  h  )  ^{2}  = 1 –  3  =  1 
160  4  4 
h  = 

=  1  
160  2 
tan 30° =  AB 
BC 
1  =  h 
3  50 
h =  50  m 
3 
tan 45° =  AB  ⇒ 


BD 
tan 30° =  AB  ⇒ 


BC 
⇒ h  3  – x = 10 
⇒ h  3  – h = 10 [From eq. (i)] 
⇒ h(  3  – 1) = 10 
⇒ h =  10  × 

= 5( 

+ 1) m  

