Examples(Set 2) - Height and Distance

6. The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is:
A. 14.8 m
B. 6.2 m
C. 12.4 m
D. 24.8 m

Answer: Option D

Explanation:
cos 60° = PS/TS
PT=2 x 12.4
= 24.8 m






7.A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it If it takes 12 minutes for the angle of depression to change from 30 degree to 45 degree how soon after this will the car reach the observation tower?
A. 14 min. 35 sec.
B. 15 min. 49 sec.
C. 16 min. 23 sec.
D. 18 min. 5 sec.

Answer: Option C

Explanation:
Let AB be the tower and C and D be the two positions of the car.
Then, ∠ACB=45∘, ∠ADB=30∘
Let, AB = h, CD = x and AC = y
AB/AC=tan45∘=1
⇒y=h
AB/AD=tan30∘=1√3
⇒h/x+y=1√3
=>x=h(√3−1)
Now, h(√3−1) is covered in 12 min.
So, h will be covered in=>
[(12/h(√3−1)) ×h]
=16min,23sec






8.A woman is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the woman's eye when at a distance of 100 metres from the tower. After 10 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water?
A. 32.42 km/hr
B. 26.28 km/hr
C. 24.22 km/hr
D. 31.25 km/hr

Answer: Option B

Explanation:
angle ACB = 45° , angle ADB = 30°, AB = 100 m
tan 45 ° = AB/ AC
=> AC = 100
tan 30 ° = AB/ AD
=> AD = 100 √ 3
CD = (AD - AC)
=100 ( √ 3 − 1 )
Required speed = Distance/ Time
= 100 ( √ 3 − 1 ) /10
=7.3 × 18/ 5 km/hr
= 26.28 km/hr






9. The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 40 m towards the tower, the angle of elevation of the top of the tower increases by 15°. The height of the tower is:
A. 62.2 m
B. 64.2 m
C. 52.2 m
D. 54.6 m

Answer: Option D

Explanation:
angle ACB = 45° , angle ADB = 30°, CD = 40 m
tan 45 ° = AB/ AC
=>AB=AC
tan 30 ° = AB/ AD
tan 30 °= AB/AC+CD
tan 30 °=AB/AB+40
AB=40/0.73
=> AB=54.6 m






10.The top of a 18 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
A. 12 metres
B. 5 metres
C. 8 metres
D. 10 metres

Answer: Option A

Explanation:
AC = 18 m, angle ADB = 30°, angle AEC = 60° ,BC = DE,BD=CE
tan 60 ° = AC/ CE
=> CE = 18/ √ 3
tan 30 ° = AB/ BD
=> 1/ √3 = AB/ (18/√3)
AB=6
BC=AC-AB
BC=DE=12 m






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