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?

Option 1: $40 \sqrt{3}$

Option 2: $\frac{151 \sqrt{3}}{6}$

Option 3: $\frac{77}{4} \sqrt{3}$

Option 4: $\frac{40 \sqrt{3}}{3}$

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?

Option 1: $\frac{120}{\sqrt{3}}$ m

Option 2: $\frac{60}{{\sqrt3}}$ m

Option 3: $\frac{50}{{\sqrt3}}$ m

Option 4: $\frac{130}{{\sqrt7}}$ m

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: $340(3 - \sqrt3)$

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 tower?

Option 1: $90$

Option 2: $45\sqrt{3}$

Option 3: $45(\sqrt{3}+1)$

Option 4: $45(\sqrt{3}+2)$

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$\sqrt{3}=1.732$)

Option 1: 25.98

Option 2: 26.58

Option 3: 27.58

Option 4: 27.98