13W O SRS I सका: ) heus a2 मकर कर्यीन धर|

1.	13W O SRS I सका: ) heus a2 मकर कर्यीन धर\| bhis pRM MR R eSe D Epy 2 WG[ e W hIB S Sh M B dnge P Mk 9T B4N
Answer» Given AB is the tower. P and Q are the points at distance of 4m and 9m respectively. From fig, PB = 4m, QB = 9m. Let angle of elevation from P be α and angle of elevation from Q be β. Given that α and β are supplementary. Thus, α + β = 90 In triangle ABP, tan α = AB/BP – (i) In triangle ABQ, tan β = AB/BQ tan (90 – α) = AB/BQ (Since, α + β = 90) cot α = AB/BQ 1/tan α = AB/BQ So, tan α = BQ/AB – (ii) From (i) and (ii) AB/BP = BQ/AB AB^2 = BQ x BP AB^2 = 4 x 9 AB^2 = 36 Therefore, AB = 6. Hence, height of tower is 6m.

13W O SRS I सका: ) heus a2 मकर कर्यीन धर| bhis pRM MR R eSe D Epy 2 WG[ e W hIB S Sh M B dnge P Mk 9T B4N

Answer»

Given AB is the tower.

P and Q are the points at distance of 4m and 9m respectively.

From fig, PB = 4m, QB = 9m.

Let angle of elevation from P be α and angle of elevation from Q be β.

Given that α and β are supplementary. Thus, α + β = 90

In triangle ABP,

tan α = AB/BP – (i)

In triangle ABQ,

tan β = AB/BQ

tan (90 – α) = AB/BQ (Since, α + β = 90)

cot α = AB/BQ

1/tan α = AB/BQ

So, tan α = BQ/AB – (ii)

From (i) and (ii)

AB/BP = BQ/AB

AB^2 = BQ x BP

AB^2 = 4 x 9

AB^2 = 36

Therefore, AB = 6.

Hence, height of tower is 6m.