if 10 theta = 1/ root 7
show that cosec ² theta - sec ²theta
upon cosec²theta + sec²theta =3/4
Answers
♧Answer♧
• tanθ = 1/_/7
= cotθ = _/7
• sec²θ = (1+tan²θ) = 1+1/7 = 8/7
• cosec²θ = (1+cot²θ) = 1+7=8
cosec²θ+sec²θ/cosec²θ−sec²θ
= 8 - 8/7/8+8/7
= 48/64
= 3/4
Proved!
Answer:
Given :
Bottom distance = 100 m.
Angle of Elevation \bf{\angle_{1}}∠
1
= 60°
Angle of Elevation \bf{\angle_{2}}∠
2
= 30°.
To find :
Height of the two poles.
Distance of the points from the feet of the poles.
Solution :
Let the height of both the poles will be h m.
Let the distance from point A and B be x m.
Hence according to the Question , the distance from point B will be (100 - x) m.
Height of the tower :
To find the height of pole (in terms of h) with respect to angle 60°.
Using tan θ and substituting the values in it, we get :
\begin{gathered}:\implies \bf{tan\:\theta = \dfrac{P}{B}} \\ \\ \\\end{gathered}
:⟹tanθ=
B
P
\begin{gathered}:\implies \bf{tan\:60^{\circ} = \dfrac{h}{x}} \\ \\ \\\end{gathered}
:⟹tan60
∘
=
x
h
\begin{gathered}:\implies \bf{\sqrt{3} = \dfrac{h}{x}}\quad[\because \bf{tan\:60^{\circ} = \sqrt{3}}] \\ \\ \\\end{gathered}
:⟹
3
=
x
h
[∵tan60
∘
=
3
]
\begin{gathered}:\implies \bf{x = \dfrac{h}{\sqrt{3}}} \\ \\ \\\end{gathered}
:⟹x=
3
h
\begin{gathered}\boxed{\therefore \bf{x = \dfrac{h}{\sqrt{3}}}} \quad \quad Eq..(i) \ \\ \\ \\\end{gathered}
∴x=
3
h
Eq..(i)
Hence the distance between base of A and B (in terms of h) is √3/h
Now , by using the tan θ and substituting the values in it, we get :
\begin{gathered}:\implies \bf{tan\:\theta = \dfrac{P}{B}} \\ \\ \\\end{gathered}
:⟹tanθ=
B
P
\begin{gathered}:\implies \bf{tan\:30^{\circ} = \dfrac{h}{100 - x}} \\ \\ \\\end{gathered}
:⟹tan30
∘
=
100−x
h
\begin{gathered}:\implies \bf{\dfrac{1}{\sqrt{3}} = \dfrac{h}{100 - x}}\quad[\because \bf{tan\:30^{\circ} = \dfrac{1}{\sqrt{3}}}] \\ \\ \\\end{gathered}
:⟹
3
1
=
100−x
h
[∵tan30
∘
=
3
1
]
\begin{gathered}:\implies \bf{\dfrac{100 - x}{\sqrt{3}} = h} \\ \\ \\\end{gathered}
:⟹
3
100−x
=h
\begin{gathered}:\implies \bf{\dfrac{100 - x}{\sqrt{3}} = h} \\ \\ \\\end{gathered}
:⟹
3
100−x
=h
\begin{gathered}:\implies \bf{100 - x = h\sqrt{3}} \\ \\ \\\end{gathered}
:⟹100−x=h
3
Now , by substituting the value of x from equation (i) , we get :
\begin{gathered}:\implies \bf{100 - \dfrac{h}{\sqrt{3}} = h\sqrt{3}} \\ \\ \\\end{gathered}
:⟹100−
3
h
=h
3
\begin{gathered}:\implies \bf{\dfrac{100\sqrt{3} - h}{\sqrt{3}} = h\sqrt{3}} \\ \\ \\\end{gathered}
:⟹
3
100
3
−h
=h
3
\begin{gathered}:\implies \bf{100\sqrt{3} - h = h\sqrt{3} \times \sqrt{3}} \\ \\ \\\end{gathered}
:⟹100
3
−h=h
3
×
3
\begin{gathered}:\implies \bf{100\sqrt{3} - h = 3h} \\ \\ \\\end{gathered}
:⟹100
3
−h=3h
\begin{gathered}:\implies \bf{100\sqrt{3} = h + 3h} \\ \\ \\\end{gathered}
:⟹100
3
=h+3h
\begin{gathered}:\implies \bf{100\sqrt{3} = 4h} \\ \\ \\\end{gathered}
:⟹100
3
=4h
\begin{gathered}:\implies \bf{\dfrac{100\sqrt{3}}{4} = h} \\ \\ \\\end{gathered}
:⟹
4
100
3
=h
\begin{gathered}:\implies \bf{25\sqrt{3} = h} \\ \\ \\\end{gathered}
:⟹25
3
=h
\begin{gathered}\boxed{\therefore \bf{h = 25\sqrt{3}}} \\ \\ \\\end{gathered}
∴h=25
3
Hence the Height of two towers is 25√3 m.
Distance from the points :
Distance between A and B :
Since, we have taken the base distance as x and we know the value of x in terms of h i.e,
\begin{gathered}:\implies \bf{x = \dfrac{h}{\sqrt{3}}} \\ \\ \\\end{gathered}
:⟹x=
3
h
Now, putting the value of h in the above equation , we get :
\begin{gathered}:\implies \bf{x = \dfrac{25\sqrt{3}}{\sqrt{3}}} \\ \\ \\\end{gathered}
:⟹x=
3
25
3
\begin{gathered}:\implies \bf{x = 25 m} \\ \\ \\\end{gathered}
:⟹x=25m
\begin{gathered}\boxed{\therefore \bf{x = 25\:m}} \\ \\ \\\end{gathered}
∴x=25m
Hence, the base distance from A to B is 25.
Distance between B and C :
We know that the distance between B and C is (100 - x) m.
So by putting the value of x in it , we get :
\begin{gathered}:\implies \bf{100 - x} \\ \\\end{gathered}
:⟹100−x
\begin{gathered}:\implies \bf{100 - 25} \\ \\\end{gathered}
:⟹100−25
\begin{gathered}:\implies \bf{75} \\ \\\end{gathered}
:⟹75
\begin{gathered}\boxed{\therefore \bf{75\:m}} \\ \\ \\\end{gathered}
∴75m