Question : The angle of elevation of the top of a tower from the top of a building whose height is 680 m is $45^{\circ}$ and the angle of elevation of the top of the same tower from the foot of the same building is $60^{\circ}$. What is the height (in m) of the tower?
Option 1: $340(3 + \sqrt3)$
Option 2: $310(3 - \sqrt3)$
Option 3: $310(3 + \sqrt3)$
Option 4: $340(3 - \sqrt3)$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $340(3 + \sqrt3)$
Solution : Let AB be the building and CD be the tower. AB = DE = 680 m $\tan 45^{\circ}$ = $\frac{CE}{AE}$ ⇒ 1 = $\frac{CE}{BD}$ ⇒ CE = BD ------------(i) $\tan 60^{\circ}$ = $\frac{CD}{BD}$ ⇒ $\sqrt3$ = $\frac{CE+DE}{CE}$ ⇒ $\sqrt3$ × CE = CE + DE ⇒ CE ($\sqrt3-1$) = 680 ⇒ CE = $\frac{680}{\sqrt3-1}$ ⇒ CE = $\frac{680(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}$ ⇒ CE = $340(\sqrt3+1)$ CD = DE + CE = $680 + 340(\sqrt3+1)$ = $340(\sqrt3+3)$ Hence, the correct answer is $340(\sqrt3+3)$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : From 40 metres away from the foot of a tower, the angle of elevation of the top of the tower is 60°. What is the height of the tower?
Question : From the top of a 20 metres high building, the angle of elevation from the top of a tower is 60° and the angle of depression of its foot is at 45°, then the height of the tower is: $(\sqrt{3} = 1.732)$
Question : A tower is broken at a point P above the ground. The top of the tower makes an angle of $60^\circ$ with the ground at Q. From another point R on the opposite side Q angle of elevation of point P is $30^\circ$. If QR = 180 m, what is the total height (in meters) of the
Question : A 1.6 m tall observer is 45 metres away from a tower. The angle of elevation from his eye to the top of the tower is 30°, then the height of the tower in metres is: (Take$\sqrt{3}=1.732$)
Question : From the top of an upright pole 17.75 m high, the angle of elevation of the top of an upright tower was 60°. If the tower was 57.75 m tall, how far away (in m) from the foot of the pole was the foot of the tower?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile