Question

plss ans. these questions ​

Answer 1

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Question

Prove that :

tan θ + tan(90 - θ) = sec θ sec(90 - θ)

Solution

Taking LHS

⇨ LHS = tan θ + tan (90 - θ)

Using trigonometric ratio

cot A = tan (90 - A)

⇨ LHS = tan θ + cot θ

⇨ LHS = $\sf{\frac{sin\theta}{cos\theta}+\frac{cos\theta}{sin\theta}}$

Taking LCM

⇨ LHS = $\sf{\frac{sin^2\theta+cos^2\theta}{sin\theta\:cos\theta}}$

Using trigonometric identity

sin²A + cos²A = 1

⇨ LHS = $\sf{\frac{1}{sin\theta\;cos\theta}}$

⇨ LHS =  cosecθ secθ

Using trigonometric ratio

cosec A = sec ( 90 - A )

⇨ LHS = sec ( 90 - θ ) sec θ = RHS

Proved .

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Question

Find the angle of elevation of the sun when shadow of a pole h metres high is √3 h metres long.

Solution

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,8)\linethickness{0.5mm}\qbezier(1,1)(3,1)(4,1)\qbezier(4,1)(4,2)(4,3)\qbezier(1,1)(3,2.3)(4,3)\put(1.5,1.12){ \theta }\put(0.7,1){ A }\put(4,0.7){ B }\put(4.2,3.2){ C }\put(4.2,2){ h }\put(2.5,0.3){ \sqrt3 h }\end{picture}$

In the figure , BC is the pole with height h , AB is the shadow of pole of length √3 h

We have to find the angle of elevation of sun that is ∠ θ

→ As we know ,

$\sf{tan\theta=\frac{opposite\:side}{adjacent\:side}}$

so,

$\sf{tan\;\theta=\frac{h}{\sqrt3\;h}}\\\\\sf{tan\;\theta=\frac{1}{\sqrt3}}$

and we know

$\sf{tan\;30^{\circ} =\frac{1}{\sqrt3}}$

Hence,

$\boxed{\sf{\theta=30^{\circ}}}$

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Question

If √3 tan θ = 1 , then find the value of sin²θ - cos²θ.

Solution

As given

$\rightarrow\sf{\sqrt 3\; tan\theta=1}$

$\rightarrow\sf{tan\theta=\frac{1}{\sqrt3}}$

Also,

$\rightarrow\sf{tan\;30^{\circ}=\frac{1}{\sqrt3}}$

Hence , we found

$\implies\underline{\underline{\sf{\theta=30^{\circ}}}}$

We have to find

$\longrightarrow\sf{sin^2\theta-cos^2\theta}$

Using, $\sf{sin\;30^{\circ}=\frac{1}{2}\;and\;cos\;30^{\circ}=\frac{\sqrt3}{2}}$

$\longrightarrow\sf{(\frac{1}{2})^2-(\frac{\sqrt3}{2})^2}$

$\longrightarrow\sf{\frac{1}{4}-\frac{3}{4}=\frac{1-3}{4}=\frac{-2}{4}=\frac{-1}{2}}$

so,

$\longrightarrow\boxed{\sf{sin^2\theta-cos^2\theta=\frac{-1}{2}}}$

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Question

A ladder 15 metres long just reaches the top of a vertical wall . If the ladder makes an angle of 60° with the wall , find the height of the wall.

Solution

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,8)\linethickness{0.5mm} \qbezier(1.5,1)(2,1)(4,1)\qbezier(4,1)(4,2)(4,4)\qbezier(1.5,1)(2,1.7)(4,4)\put(1.9,1,2){ 60^{\circ} }\put(4.2,2.5){ x }\put(2.1,2.6){ 15\;m }\put(0.9,1){ P }\put(4.3,1){ Q } \put(4,4.2){ R }\end{picture}$

In the figure ,

PR is the ladder , RQ is the Wall with height x

We have to find the height of wall

so,

$\implies\sf{sin\:60^{\circ}=\frac{RQ}{PR}}\\\\\implies\sf{\frac{\sqrt3}{2}=\frac{x}{15}}\\\\\implies\boxed{\sf{x=\frac{15\;\sqrt3}{2}\:m}}$

Hence , height of wall is $\sf{\frac{15\;\sqrt3}{2}\;m}$ .

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Answer 2

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$\large{\red{\underline{\tt{Question:\:-}}}}$

tan θ + tan(90 - θ) = sec θ sec(90 - θ)

$\large{\red{\underline{\tt{Solution\::-}}}}$

$\dashrightarrow$ $\sf tan \theta + tan ( 90 - \theta )$

$\dashrightarrow$ $\sf tan \theta + cot \theta$

$\dashrightarrow$ $\sf \dfrac{sin \theta}{cos \theta} + \dfrac{cos \theta}{sin \theta}$

$\dashrightarrow$ $\sf \dfrac{sin^{2} \theta + cos^{2} \theta}{sin \theta \times cos \theta}$

$\dashrightarrow$ $\sf \dfrac{1}{sin \theta \times cos \theta}$

$\dashrightarrow$ $\sf Cosec \theta \times Sec \theta$

$\dashrightarrow$ $\sf Sec (90 - \theta) \times Sec \theta$

$\orange\dashrightarrow$ $\underline{\boxed{\orange{\sf sec (90 - \theta) = Cos \theta)}}}$ $\dagger$

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$\large{\purple{\underline{\tt{Question\::-}}}}$

Find the angle of elevation of the sun when shadow of a pole h metres high is √3 h metres long.

$\large{\purple{\underline{\tt{figure\::-}}}}$

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,8)\linethickness{0.5mm}\qbezier(1,1)(3,1)(4,1)\qbezier(4,1)(4,2)(4,3)\qbezier(1,1)(3,2.3)(4,3)\put(1.5,1.12){ \theta }\put(0.7,1){ A }\put(4,0.7){ B }\put(4.2,3.2){ C }\put(4.2,2){ h }\put(2.5,0.3){ \sqrt3 h }\end{picture}$

$\large{\purple{\underline{\tt{Solution\::-}}}}$

$\sf \triangle ABC$

$\longrightarrow$ $\sf tan \theta = \dfrac{CB}{AB} = \dfrac{h}{\sqrt{3}h}$

$\longrightarrow$ $\sf tan \theta = \dfrac{1}{\sqrt 3}$

$\longrightarrow$ $\sf tan\theta = 30^{\circ}$

$\longrightarrow$ $\sf tan \theta = tan\:30^{\circ}$

$\pink\longrightarrow$ $\underline{\boxed{\pink{\sf \theta = 30^{\circ}}}}$ $\dagger$

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$\large{\blue{\underline{\tt{Question \::-}}}}$

If √3 tan θ = 1 , then find the value of sin²θ - cos²θ.

$\large{\blue{\underline{\tt{Solution\::-}}}}$

$\leadsto$ $\sf \sqrt{3} tan \theta = 1$

$\leadsto$ $\sf tan \theta = \dfrac{1}{3}$

$\leadsto$ $\sf tan \theta = 30^{\circ}$

$\leadsto$ $\sf sin^{2} \theta - cos^{2} \theta$

$\leadsto$ $\sf sin^{2} (30^{\circ}) - cos^{2} (30^{\circ})$

$\leadsto$ $\sf \left (\dfrac{1}{2}\right)^{2} - \left(\dfrac{\sqrt 3}{2}\right)^{2}$

$\leadsto$ $\sf \dfrac{1}{4} - \dfrac{3}{4}$

$\green\leadsto$ $\large\underline{\boxed{\green{\sf \dfrac{-1}{2}}}}$ $\dagger$

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$\large{\gray{\underline{\tt{Question\::-}}}}$

A ladder 5m long, leans against a wall so that it makes an angle of 60° with the horizontal ground. calculate how far up the wall the ladder reaches.

$\large{\gray{\underline{\tt{Figure\::-}}}}$

$\setlength{\unitlength}{1.6cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{C}}\put(10.6,1){\large\sf{B}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.4,2){\sf{\large{x}}}\put(9.3,2){\sf{\large{15m}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\put(9.4,1.2){\sf\large{60^{\circ} }}\end{picture}$

$\large{\gray{\underline{\tt{Solution\::-}}}}$

$\mapsto$ $\sf sin (C) = \dfrac{Perpendicular}{Hypotenuse}$

$\mapsto$ $\sf sin (60^{\circ}) = \dfrac{AC}{AB}$

$\mapsto$ $\sf \dfrac{\sqrt 3}{2} = \dfrac{AC}{5m}$

$\purple\mapsto$ $\underline{\boxed{\purple{\sf AC = \dfrac{15 \sqrt{3m}}{2}}}}$ $\dagger$

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Extra shots :-

$\dagger\:\sf Trigonometric\:Values :\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D \hat{e} fined\end{tabular}}$

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