The owner of a boat must depart from point $P$ and arrive at point $R$ by means of two linear displacements and navigating at a constant speed. This trip will be made during the night, and since he has only a compass and a clock, he planned his route as follows: $1^{\text{st}}$ - depart from point $P$ in direction 110 and navigate for 4 hours, reaching a point $Q$; $2^{\text{nd}}$ - depart from point $Q$ in direction 90 and navigate for 2 hours, reaching the destination point $R$.
However, when directing the boat for the first displacement, he did so in direction 340, instead of 110. With this, he made the following displacements: $1^{\text{st}}$ - departed from point $P$ in direction 340 and navigated for 4 hours, reaching a point $S$; $2^{\text{nd}}$ - departed from point $S$ in direction 90 and navigated for 2 hours, reaching point $T$.
The boat owner only realized the mistake upon arriving at point $T$. With this, he now needs to define the direction and navigation time that will allow him, departing from point $T$, to reach the destination point $R$ through a straight route.
Consider 0.64 as an approximation for $\cos 50°$. The direction and approximate navigation time that the boat owner should use are, respectively,
(A) 135 and 7 hours and 15 minutes.
(B) 45 and 7 hours and 15 minutes.
(C) 135 and 12 hours.
(D) 135 and 6 hours.
(E) 45 and 6 hours.

15. As shown in the figure, a car is traveling due west on a horizontal road. At point A, the mountain peak D on the north side of the road is measured to be in the direction of $30 ^ { 0 }$ west of north. After traveling 600 m to reach point B, the mountain peak is measured to be in the direction of $75 ^ { 0 }$ west of north, with an angle of elevation of $30 ^ { 0 }$. Then the height of the mountain $\mathrm { CD } =$ $\_\_\_\_$ m. [Figure]

13. As shown in the figure, a car is traveling due west on a horizontal road. At point $A$, the mountain peak $D$ on the north side of the road is measured to be in the direction $30°$ west of north. After traveling 600 m to reach point $B$, the peak is measured to be in the direction $75°$ west of north with an elevation angle of $30°$. The height of the mountain $CD = $ $\_\_\_\_$ m.
[Figure]
Figure for Question 13
[Figure]
Figure for Question 14

8. On December 8, 2020, China and Nepal jointly announced that the latest height of Mount Everest is 8848.86 m. The trigonometric height measurement method is one of the methods for measuring the height of Mount Everest. The figure on the right is a schematic diagram of the trigonometric height measurement method. There are three points $A, B, C$, and their projections $A', B', C'$ on the same horizontal plane satisfy $\angle A'C'B' = 45°$, $\angle A'B'C' = 60°$. The angle of elevation from point $C$ to point $B$ is $15°$. The difference between $BB'$ and $CC'$ is 100 m. The angle of elevation from point $B$ to point $A$ is $45°$. Then the height difference between points $A$ and $C$ to the horizontal plane $A'B'C'$ is $AA' - CC'$ approximately equals ($\sqrt{3} \approx 1.732$)
A. 346
B. 373
C. 446
D. 473

An aeroplane $P$ is moving in the air along a straight line path which passes through the points $P_1$ and $P_2$, and makes an angle $\alpha$ with the ground. Let $O$ be the position of an observer. When the plane is at the position $P_1$ its angle of elevation is $30^\circ$ and when it is at $P_2$ its angle of elevation is $60^\circ$ from the position of the observer. Moreover, the distances of the observer from the points $P_1$ and $P_2$ respectively are 100 metres and $500/3$ metres.
Then $\alpha$ is equal to
(a) $\tan^{-1}\{(2-\sqrt{3})/(2\sqrt{3}-1)\}$
(b) $\tan^{-1}\{(2\sqrt{3}-3)/(4-2\sqrt{3})\}$
(c) $\tan^{-1}\{(2\sqrt{3}-2)/(5-\sqrt{3})\}$
(d) $\tan^{-1}\{(6-\sqrt{3})/(6\sqrt{3}-1)\}$

Two poles, $AB$ of length two metres and $CD$ of length twenty metres are erected vertically with bases at $B$ and $D$. The two poles are at a distance not less than twenty metres. It is observed that $\tan \angle ACB = 2/77$. The distance between the two poles is
(A) $72 m$
(B) 68 m
(C) 24 m
(D) 24.27 m

Two poles, $AB$ of length two metres and $CD$ of length twenty metres are erected vertically with bases at $B$ and $D$. The two poles are at a distance not less than twenty metres. It is observed that $\tan \angle ACB = 2/77$. The distance between the two poles is
(A) $72 m$
(B) 68 m
(C) 24 m
(D) 24.27 m

Two poles, $A B$ of length two metres and $C D$ of length twenty metres are erected vertically with bases at $B$ and $D$. The two poles are at a distance not less than twenty metres. It is observed that $\tan \angle A C B = 2 / 77$. The distance between the two poles is
(A) $72 m$
(B) 68 m
(C) 24 m
(D) 24.27 m

A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point $O$ on the ground is $45 ^ { \circ }$. It flies off horizontally straight away from the point $O$. After one second, the elevation of the bird from $O$ is reduced to $30 ^ { \circ }$. Then the speed (in m/s) of the bird is
(1) $20 \sqrt { 2 }$
(2) $20 ( \sqrt { 3 } - 1 )$
(3) $40 ( \sqrt { 2 } - 1 )$
(4) $40 ( \sqrt { 3 } - \sqrt { 2 } )$

If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot of the tower, are $30^\circ$, $45^\circ$ and $60^\circ$ respectively, then the ratio $AB : BC$, is:
(1) $\sqrt{3} : 1$
(2) $\sqrt{3} : \sqrt{2}$
(3) $1 : \sqrt{3}$
(4) $2 : 3$

If the angles of elevation of the top of a tower from three collinear points $A , B$ and $C$ on a line leading to the foot of the tower are $30 ^ { \circ } , 45 ^ { \circ }$ and $60 ^ { \circ }$ respectively, then the ratio $AB : BC$, is
(1) $2 : 3$
(2) $\sqrt { 3 } : 1$
(3) $\sqrt { 3 } : \sqrt { 2 }$
(4) $1 : \sqrt { 3 }$

The angle of elevation of the top of a vertical tower from a point A , due east of it is $45 ^ { \circ }$. The angle of elevation of the top of the same tower from a point B , due south of A is $30 ^ { \circ }$. If the distance between A and B is $54 \sqrt { 2 } m$, then the height of the tower (in meters), is:
(1) 108
(2) $36 \sqrt { 3 }$
(3) $54 \sqrt { 3 }$
(4) 54

Let a vertical tower $AB$ have its end $A$ on the level ground. Let $C$ be the mid-point of $AB$ and $P$ be a point on the ground such that $AP = 2AB$. If $\angle BPC = \beta$, then $\tan\beta$ is equal to:
(1) $\dfrac{6}{7}$
(2) $\dfrac{1}{4}$
(3) $\dfrac{2}{9}$
(4) $\dfrac{4}{9}$

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min for the angle of depression of the car to change from $30 ^ { \circ }$ to $45 ^ { \circ }$, then the time taken (in $\min$ ) by the car to reach the foot of the tower is
(1) $\frac { 9 } { 2 } ( \sqrt { 3 } + 1 )$
(2) $9 ( \sqrt { 3 } + 1 )$
(3) $18 ( \sqrt { 3 } - 1 )$
(4) $9 ( \sqrt { 3 } - 1 )$

$P Q R$ is a triangular park with $P Q = P R = 200 \mathrm {~m}$. A T.V. tower stands at the mid-point of $Q R$. If the angles of elevation of the top of the tower at $P , Q$ and $R$ are respectively, $45 ^ { \circ } , 30 ^ { \circ }$ and $30 ^ { \circ }$, then the height of the tower (in m) is:
(1) $50 \sqrt { 2 }$
(2) 100
(3) 50
(4) $100 \sqrt { 3 }$

Consider a triangular plot $A B C$ with sides $A B = 7 m , B C = 5 m$ and $C A = 6 m$. A vertical lamp-post at the mid-point $D$ of $A C$ subtends an angle $30 ^ { \circ }$ at $B$. The height (in $m$) of the lamp-post is:
(1) $2 \sqrt { 21 }$
(2) $\frac { 2 } { 3 } \sqrt { 21 }$
(3) $\frac { 3 } { 2 } \sqrt { 21 }$
(4) $7 \sqrt { 3 }$

Two poles standing on a horizontal ground are of heights $5 m$ and $10 m$ respectively. The line joining their tops makes an angle of $15 ^ { \circ }$ with the ground. Then the distance (in $m$) between the poles, is
(1) $10 ( \sqrt { 3 } - 1 )$
(2) $\frac { 5 } { 2 } ( 2 + \sqrt { 3 } )$
(3) $5 ( 2 + \sqrt { 3 } )$
(4) $5 ( \sqrt { 3 } + 1 )$

The angle of elevation of a cloud $C$ from a point $P$, $200$ m above a still lake is $30 ^ { \circ }$. If the angle of depression of the image of $C$ in the lake from the point $P$ is $60 ^ { \circ }$, then $PC$ (in m) is equal to
(1) 100
(2) $200 \sqrt { 3 }$
(3) 400
(4) $400 \sqrt { 3 }$

The angle of elevation of the summit of a mountain from a point on the ground is $45^{\circ}$. After climbing up one km towards the summit at an inclination of $30^{\circ}$ from the ground, the angle of elevation of the summit is found to be $60^{\circ}$. Then the height (in km) of the summit from the ground is:
(1) $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
(2) $\frac{\sqrt{3}+1}{\sqrt{3}-1}$
(3) $\frac{1}{\sqrt{3}-1}$
(4) $\frac{1}{\sqrt{3}+1}$

The angle of elevation of a jet plane from a point $A$ on the ground is $60 ^ { \circ }$. After a flight of 20 seconds at the speed of 432 km / hour, the angle of elevation changes to $30 ^ { \circ }$. If the jet plane is flying at a constant height, then its height is:
(1) $1200 \sqrt { 3 } \mathrm {~m}$
(2) $2400 \sqrt { 3 } \mathrm {~m}$
(3) $1800 \sqrt { 3 } \mathrm {~m}$
(4) $3600 \sqrt { 3 } \mathrm {~m}$

Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:
(1) 25
(2) 30
(3) $20 \sqrt { 3 }$
(4) $25 \sqrt { 3 }$

A pole stands vertically inside a triangular park $ABC$. Let the angle of elevation of the top of the pole from each corner of the park be $\frac { \pi } { 3 }$. If the radius of the circumcircle of $\triangle ABC$ is 2 , then the height of the pole is equal to:
(1) $\frac { 2 \sqrt { 3 } } { 3 }$
(2) $2 \sqrt { 3 }$
(3) $\sqrt { 3 }$
(4) $\frac { 1 } { \sqrt { 3 } }$

A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point $A$, with uniform speed. At that point, angle of depression of the boat with the man's eye is $30 ^ { \circ }$ (Ignore man's height). After sailing for 20 seconds, towards the base of the tower (which is at the level of water), the boat has reached a point $B$, where the angle of depression is $45 ^ { \circ }$. Then the time taken (in seconds) by the boat from $B$ to reach the base of the tower is :
(1) 10
(2) $10 ( \sqrt { 3 } - 1 )$
(3) $10 \sqrt { 3 }$
(4) $10 ( \sqrt { 3 } + 1 )$

A vertical pole fixed to the horizontal ground is divided in the ratio 3 : 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is :
(1) $8 \sqrt { 10 }$
(2) $6 \sqrt { 10 }$
(3) $12 \sqrt { 10 }$
(4) $12 \sqrt { 15 }$

The angle of elevation of the top of a tower from a point $A$ due north of it is $\alpha$ and from a point $B$ at a distance of 9 units due west of $A$ is $\cos ^ { - 1 } \left( \frac { 3 } { \sqrt { 13 } } \right)$. If the distance of the point $B$ from the tower is 15 units, then $\cot \alpha$ is equal to
(1) $\frac { 6 } { 5 }$
(2) $\frac { 9 } { 5 }$
(3) $\frac { 4 } { 3 }$
(4) $\frac { 7 } { 3 }$