Aptitude - Height and Distance - Discussion
Discussion Forum : Height and Distance - General Questions (Q.No. 1)
1.
Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:
Answer: Option
Explanation:
Let AB be the lighthouse and C and D be the positions of the ships.
Then, AB = 100 m, 
ACB = 30° and ADB = 45°.
| AB | = tan 30° = | 1 |  AC = AB x 3 = 1003 m. |
| AC | 3 |
| AB | = tan 45° = 1 AD = AB = 100 m. |
| AD |
 CD = (AC + AD) |
= (1003 + 100) m |
| = 100(3 + 1) | |
| = (100 x 2.73) m | |
| = 273 m. |
Discussion:
104 comments Page 1 of 11.
Bheemu said:
2 years ago
why tan 30°? Please explain me.
(13)
SAI ESVAR said:
3 years ago
Can anyone explain this problem clearly to understand?
(11)
Akshay said:
2 years ago
Short tricks
(1)30_60_90
1:√3:2.
So,
∆adb
a = 1×100.
= 100m.
b = √3×100.
= 173m.
d = 100 × 2
= 200m.
(2) 45_45_90
1:1:√2.
∆dcb
C = 1×100.
= 100m.
B = 1×100,
= 100m.
D = √2×100.
= 141m.
Total
ADC = 173 + 100,
= 273m.
(1)30_60_90
1:√3:2.
So,
∆adb
a = 1×100.
= 100m.
b = √3×100.
= 173m.
d = 100 × 2
= 200m.
(2) 45_45_90
1:1:√2.
∆dcb
C = 1×100.
= 100m.
B = 1×100,
= 100m.
D = √2×100.
= 141m.
Total
ADC = 173 + 100,
= 273m.
(11)
HECKER said:
3 years ago
Yes, right @Kumar.
(6)
Khaleeq ur rehman said:
4 years ago
According to My observation, when the angle of elevation is 45 degree then base and perpendicular must be equal which is 100 according to AD measurement if another elevation is less then from another which is 30 degree of AC then how is possible that distance/length of AC increase from another distance. Please anyone explain.
(5)
Hemanth said:
3 years ago
Why we use the tan formula? Anyone explain me.
(5)
Chand said:
1 year ago
Sin30 = 100/B.
1/2 = 100/B.
B = 200.
Sin45 = 100/B.
1/√2 = 100/B,
B = 100×√2,
B = 100×1.41.
B = 141,
B+B = 200 + 141,
= 341.
1/2 = 100/B.
B = 200.
Sin45 = 100/B.
1/√2 = 100/B,
B = 100×√2,
B = 100×1.41.
B = 141,
B+B = 200 + 141,
= 341.
(4)
Yashish yenduva said:
1 year ago
By using tan we can easily get solution.
(4)
Yeshi samdup said:
5 years ago
Sine=opp/hyp.
Cos=adj/hyp..
Tan=opp/adj.
Tan30 = 100/CA.
CA = 100/tan30,
CA = 173.
Tan45 = 100/AD.
AD = 100/tan45,
AD = 100.
Therefore, CA+AD.
173+100.
=273.
Cos=adj/hyp..
Tan=opp/adj.
Tan30 = 100/CA.
CA = 100/tan30,
CA = 173.
Tan45 = 100/AD.
AD = 100/tan45,
AD = 100.
Therefore, CA+AD.
173+100.
=273.
(3)
Anudeep said:
4 years ago
Why we are taking √3? Explain please.
(3)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers