The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
NCERT Class X
Mathematics - Mathematics
Chapter _SOME APPLICATIONS OF
TRIGONOMETRY
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Diagram is in attachment below
Let AB = h meters.
therefore, In right ΔABC, we have:
AB/AC = tan θ
⇒ h/9 = tan θ .......(1)
In right ΔABD, we have:
AB/AD = tan (90°- θ) = cot θ
⇒ h/4 = cot θ .......(2)
Multiplying (1) and (2), we get
h/9 × h/4 = tan θ ×cot θ = 1 {tan θ X cot θ =1}
⇒ h²/36 = 1 ⇒ h² = 36
⇒ h = ± 6 m
⇒ h = 6 m {height is positive only}
Thus, the height of the tower is 6 m.
MARK IT AS BEST
Let AB = h meters.
therefore, In right ΔABC, we have:
AB/AC = tan θ
⇒ h/9 = tan θ .......(1)
In right ΔABD, we have:
AB/AD = tan (90°- θ) = cot θ
⇒ h/4 = cot θ .......(2)
Multiplying (1) and (2), we get
h/9 × h/4 = tan θ ×cot θ = 1 {tan θ X cot θ =1}
⇒ h²/36 = 1 ⇒ h² = 36
⇒ h = ± 6 m
⇒ h = 6 m {height is positive only}
Thus, the height of the tower is 6 m.
MARK IT AS BEST
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