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samanarthak shabd
Answers
Answer:
Correction in (Q1):
In the figure, not drawn to scale, TF is a tower. The elevation of T from A is x°, where tan x = 2/5 and AF = 200m. The elevation of T from B, where AB = 80m, is y°. Calculate:
(i) The height of the tower TF.
(ii) The angle y, correct to the nearest degree.
Solution:
According to the question:
\Longrightarrow \rm tanx = \dfrac{2}{5}⟹tanx=
5
2
We know that:
\Longrightarrow \rm tanx = \dfrac{Side \ opposite \ to \ x}{Side \ adjacent \ to \ x}⟹tanx=
Side adjacent to x
Side opposite to x
\Longrightarrow \rm tanx = \dfrac{TF}{AF}⟹tanx=
AF
TF
Substitute tanx = 2/5 above.
\Longrightarrow \rm \dfrac{2}{5} = \dfrac{TF}{AF}⟹
5
2
=
AF
TF
\Longrightarrow \rm \dfrac{2}{5} = \dfrac{TF}{200}⟹
5
2
=
200
TF
\Longrightarrow \rm \dfrac{2 \times 200}{5} = TF⟹
5
2×200
=TF
\Longrightarrow \rm 2 \times 40 = TF⟹2×40=TF
\Longrightarrow \rm{TF = 80 \ meters.}⟹TF=80 meters.
∴ (i) The height of the tower is 80 meters.
In ΔTBF:
\Longrightarrow \rm tany = \dfrac{Side \ opposite \ to \ y}{Side \ adjacent \ to \ y}⟹tany=
Side adjacent to y
Side opposite to y
\Longrightarrow \rm tany = \dfrac{TF}{BF}⟹tany=
BF
TF
\Longrightarrow \rm tany = \dfrac{80}{AF - AB}⟹tany=
AF−AB
80
\Longrightarrow \rm tany = \dfrac{80}{200 - 80}⟹tany=
200−80
80
\Longrightarrow \rm tany = \dfrac{80}{120}⟹tany=
120
80
\Longrightarrow \rm tany = \dfrac{8}{12}⟹tany=
12
8
\Longrightarrow \rm tany = \dfrac{2}{3}⟹tany=
3
2
\Longrightarrow \rm y = tan^{-1} \times \dfrac{2}{3} \ degrees.⟹y=tan
−1
×
3
2
degrees.
or
\Longrightarrow \rm y \approx 34^{\circ}⟹y≈34
∘
∴ The angle y, correct to the nearest degree is 34°.
_____________________
(Q2) 'A' can do a piece of work in x days and B can do it in (x + 16) days. If both start working together, they can do it in 15 days. Find 'x'.
Solution:
\Longrightarrow \sf Work \ done \ by \ 'A' = \dfrac{1}{Time \ taken \ in \ days}⟹Work done by
′
A
′
=
Time taken in days
1
\Longrightarrow \sf Work \ done \ by \ 'A' = \dfrac{1}{x}⟹Work done by
′
A
′
=
x
1
\Longrightarrow \sf Work \ done \ by \ 'B' = \dfrac{1}{Time \ taken \ in \ days}⟹Work done by
′
B
′
=
Time taken in days
1
\Longrightarrow \sf Work \ done \ by \ 'A' = \dfrac{1}{x + 16}⟹Work done by
′
A
′
=
x+16
1
If both of them work together, they can finish the work in 15 days.
\Longrightarrow \sf Work \ done \ by \ 'A' + Work \ done \ by \ 'B' = \dfrac{1}{No: \ of \ days \ taken}⟹Work done by
′
A
′
+Work done by
′
B
′
=
No: of days taken
1
\Longrightarrow \sf \ \dfrac{1}{x} + \dfrac{1}{x + 16} = \dfrac{1}{15}⟹
x
1
+
x+16
1
=
15
1
Take LCM:
\Longrightarrow \sf \ \dfrac{x + 16 + x}{x(x + 16)} = \dfrac{1}{15}⟹
x(x+16)
x+16+x
=
15
1
\Longrightarrow \sf \ \dfrac{2x + 16}{x^2 + 16x} = \dfrac{1}{15}⟹
x
2
+16x
2x+16
=
15
1
Cross multiply:
\Longrightarrow \sf 15(2x + 16) = x^2 + 16x⟹15(2x+16)=x
2
+16x
\Longrightarrow \sf 30x + 240 = x^2 + 16x⟹30x+240=x
2
+16x
\Longrightarrow \sf x^2 + 16x - 30x - 240 = 0⟹x
2
+16x−30x−240=0
Split the middle term:
\Longrightarrow \sf x^2 - 14x - 240 = 0⟹x
2
−14x−240=0
Sum ---: -14
Product ---: -240
Split ---: 10 × -24
\Longrightarrow \sf x^2 + 10x - 24x - 240 = 0⟹x
2
+10x−24x−240=0
\Longrightarrow \sf x(x + 10) - 24(x + 10) = 0⟹x(x+10)−24(x+10)=0
\Longrightarrow \sf (x - 24) (x + 10) = 0⟹(x−24)(x+10)=0
Compare both the roots/solutions with zero.
Case I
\Longrightarrow \sf x -24 = 0⟹x−24=0
\Longrightarrow \sf x = 24⟹x=24
Therefore, x = 24.
Case II
\Longrightarrow \sf x + 10 = 0⟹x+10=0
\Longrightarrow \sf x = -10⟹x=−10
This is an invalid case as time cannot be negative. therefore x ≠ 10.
Hence the value of 'x' is 24.
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