Question

1. In A POR, if PQ=PR. and R = 70° then 2Q=?​

Answer 1

Answer:

Answer :-

θ = 60°

Explanation :-

\begin{lgathered}\rm \dfrac{cos \theta}{1 - sin \theta } + \dfrac{cos \theta}{1 + sin \theta} = 4 \\ \\ \\\end{lgathered}

1−sinθ

cosθ

+

1+sinθ

cosθ

=4

\begin{lgathered}\text{Taking LCM} \\ \\ \\\end{lgathered}

Taking LCM

\begin{lgathered}\rm \implies \dfrac{cos \theta(1 + sin \theta) + cos \theta(1 - sin \theta)}{(1 - sin \theta)(1 + sin \theta) }= 4 \\ \\ \\\end{lgathered}

⟹

(1−sinθ)(1+sinθ)

cosθ(1+sinθ)+cosθ(1−sinθ)

=4

\begin{lgathered}\rm \implies \dfrac{cos \theta + cos \theta .sin \theta+ cos \theta - cos \theta .sin \theta}{1^{2} - sin^{2} \theta }= 4 \\ \\ \\\end{lgathered}

⟹

1

2

−sin

2

θ

cosθ+cosθ.sinθ+cosθ−cosθ.sinθ

=4

\begin{lgathered}\boxed{\boxed{ \bf \because (x - y)(x + y) = {x}^{2} - {y}^{2} }} \\ \\ \\\end{lgathered}

∵(x−y)(x+y)=x

2

−y

2

\begin{lgathered}\rm \implies \dfrac{2cos \theta }{1 - sin^{2} \theta }= 4 \\ \\ \\\end{lgathered}

⟹

1−sin

2

θ

2cosθ

=4

\begin{lgathered}\rm \implies \dfrac{2cos \theta }{cos^{2} \theta}= 4 \\ \\ \\\end{lgathered}

⟹

cos

2

θ

2cosθ

=4

\begin{lgathered}\boxed{\boxed{ \bf \because 1 - sin^{2} \theta = cos^{2} \theta }} \\ \\ \\\end{lgathered}

∵1−sin

2

θ=cos

2

θ

\begin{lgathered}\rm \implies \dfrac{2cos \theta }{cos\theta.cos \theta}= 4 \\ \\ \\\end{lgathered}

⟹

cosθ.cosθ

2cosθ

=4

\begin{lgathered}\rm \implies \dfrac{2 }{cos \theta}= 4 \\ \\ \\\end{lgathered}