a distant cliff has 10°and 20° angles of elevation at two places 100m apart. what is the height of the cliff?
Answers
Answer:
Given:
A distant cliff has 10° and 12° angles of elevation at two places as 100 m apart
To find:
The height of the cliff
Solution:
From the figure attached below, we have
AD = height of the cliff
BC = distance between the two places = 100 m
∠ABC = 10° = angle of elevation from point B to the top of the cliff
∠ACB = 12° = angle of elevation from point C to the top of the cliff
Formula to be used:
Trigonometric ratio of a triangle:- tan θ =
Let's assume,
"x" meters → be the distance between B and D
"100 - x" meters → be the distance between C and D
Consider ΔABD, we have
AD = perpendicular
BD = base
θ = 10°
∴ tan 10° =
⇒ tan 10° =
⇒ 0.176x = AD ....... (i)
Consider ΔADC, we have
AD = perpendicular
CD = base
θ = 12°
∴ tan 12° =
⇒ tan 12° =
⇒ 0.212(100 - x) = AD ....... (ii)
Now, comparing eq. (i) & (ii), we get
0.176x = 0.212(100 - x)
⇒ 0.176x = 21.2 - 0.212x
⇒ 0.176x + 0.212x = 21.2
⇒ 0.388x = 21.2
⇒ x =
⇒ x = 54.63 m
By substituting the value of x in eq.(i), we get
AD = 0.176x = 0.176 * 54.63 = 9.61 m
Thus, the height of the cliff is 9.61 m.
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Answer:
Given:
A distant cliff has 10° and 12° angles of elevation at two places as 100 m apart
To find:
The height of the cliff
Solution:
From the figure attached below, we have
AD = height of the cliff
BC = distance between the two places = 100 m
∠ABC = 10° = angle of elevation from point B to the top of the cliff
∠ACB = 12° = angle of elevation from point C to the top of the cliff
Formula to be used:
Trigonometric ratio of a triangle:- tan θ = \frac{Perpendicular}{Base}
Base
Perpendicular
Let's assume,
"x" meters → be the distance between B and D
"100 - x" meters → be the distance between C and D
Consider ΔABD, we have
AD = perpendicular
BD = base
θ = 10°
∴ tan 10° = \frac{AD}{BD}
BD
AD
⇒ tan 10° = \frac{AD}{x}
x
AD
⇒ 0.176x = AD ....... (i)
Consider ΔADC, we have
AD = perpendicular
CD = base
θ = 12°
∴ tan 12° = \frac{AD}{CD}
CD
AD
⇒ tan 12° = \frac{AD}{100 - x}
100−x
AD
⇒ 0.212(100 - x) = AD ....... (ii)
Now, comparing eq. (i) & (ii), we get
0.176x = 0.212(100 - x)
⇒ 0.176x = 21.2 - 0.212x
⇒ 0.176x + 0.212x = 21.2
⇒ 0.388x = 21.2
⇒ x = \frac{21.2}{0.388}
0.388
21.2
⇒ x = 54.63 m
By substituting the value of x in eq.(i), we get
AD = 0.176x = 0.176 * 54.63 = 9.61 m
Thus, the height of the cliff is 9.61 m..
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