8. A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Answer:
Let AB be the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured.
In ∆BCD,
\(\frac{BC}{CD}= tan\,45^\circ\)
\(\frac{BC}{CD}= 1\)
BC = CD
In ∆ACD,
Therefore, the height of the pedestal is 0.8(√3 + 1) m.
9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Answer:
Let AB be the building and CD be the tower.
In ∆CDB,
\(\frac{CD}{BD}=tan\,60^\circ\)
\(\frac{50}{BD}= \sqrt{3}\)
\(BD = \frac{50}{\sqrt{3}}\)
In ∆ABD,
\(\frac{AB}{BD}= tan\,30^\circ\)
\(AB=\frac{50}{\sqrt{3}}\times\frac{1}{\sqrt{3}}=\frac{50}{3}= 16\frac{2}{3}\)
Therefore, the height of the building is 16\(\frac{2}{3}\) m.
10. Two poles of equal heights are standing opposite each other and either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30º, respectively. Find the height of poles and the distance of the point from the poles.
Answer:
Let AB and CD be the poles and O is the point from where the elevation angles are measured.
In ∆ABO,
\(\frac{AB}{BO}=tan\,60^\circ\)
\(\frac{AB}{BO}= \sqrt{3}\)
\(BO= \frac{AB}{\sqrt{3}}\)
In ∆CDO,
Since the poles are of equal heights,
CD = AB
DO = BD − BO = (80 − 20) m = 60 m
Therefore, the height of poles is 20√3 m and the point is 20 m and 60 m far from these poles.
11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.
Answer:
In ∆ABC,
\(\frac{AB}{BC}=tan\,60^\circ\)
\(\frac{AB}{BC}=\sqrt{3}\)
BC = \(\frac{AB}{\sqrt{3}}\)
In ∆ABD,
Therefore, the height of the tower is 10√3 m and the width of the canal is 10 m.
12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Answer:
Let AB be a building and CD be a cable tower.
In ∆ABD,
\(\frac{AB}{BD}=tan\,45^\circ\)
\(\frac{7}{BD}= 1\)
BD = 7 m
In ∆ACE,
AC = BD = 7 m
Therefore, the height of the cable tower is 7(√3 + 1) m.
13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Answer:
Let AB be the lighthouse and the two ships be at point C and D respectively.
In ∆ABC,
\(\frac{AB}{BC}=tan\,45^\circ\)
\(\frac{75}{BC}= 1\)
BC = 75 m
In ∆ABD,
Therefore, the distance between the two ships is 75(√3 − 1) m.
14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.
Answer:
Let the initial position A of balloon change to B after some time and CD be the girl.
In ∆ACE,
In ∆BCG,
Distance travelled by balloon = EG = CG − CE
= (87√3 - 29√3)m
= 58√3 m
