Question

Height of a Climber : Himalayan Trekking Club has just hiked to the south rim of a large canyon, when they spot a climber attempting to scale the taller northern face. Knowing the distance between the sheer walls of the northern and southern faces of the canyon is approximately 150 meter, they attempt to compute the distance remaining for the climbers to reach the top of the northern rim. Using a homemade transit, they sight an angle of depression of 60° to the bottom of the north face, and angles of elevation of 30° and 45° to the climbers and top of the northern rim respectively. 

(i) How high is the southern rim of the canyon? 

(ii) How high is the northern rim? 

(iii) How much farther until the climber reaches the top?

Answer 1

Height of a climber question

Answer 2

\(DF = CE = 150 m\)

In \(\triangle ACE,\)

\(\tan45° = \frac{AC}{CE}\)

⇒ \(AC = CE\)

⇒ \(AC = 150 m\)

In \(\triangle CED,\)

\(\tan60° = \frac{CD}{CE}\)

⇒ \(CD = \sqrt 3CE\)

\(= \sqrt 3 \times 150 m\)

(i) \(\therefore \) Height of southern rim of the canyon 

\(= EF = CD = 150\sqrt 3\)

(ii) Height of the northern rim of the canyon 

\(= AD = AC + CD = (150 + 150\sqrt 3)m = 150(\sqrt 3 +1)m.\)

(iii) In \(\triangle BCE,\) 

\(\tan 30° = \frac{BC}{CE}\)

⇒ \(BC = \frac 1{\sqrt 3} CE = \frac{150} {\sqrt 3} m\)

\(\therefore AB = AC - BC \)

\(= (150 - \frac{150} {\sqrt3})m\)

\(= 150 \frac{(\sqrt 3 - 1)}{\sqrt 3} m\)

\(= 50(3 - \sqrt 3)m\)

Hence, climber needs to climb 50(3 - √3)m distance more to reach the top.