Question

if 10 theta = 1/ root 7
show that cosec ² theta - sec ²theta
upon cosec²theta + sec²theta =3/4​

Answer 1

♧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 2

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