Question

pls solve
trigonometry​

Answer 1

Answer:

1. PQ/PR
2. QR/PR
3. PR^2
4. PR^2/PR^2=1
5. for last answer in given picture

Answer 2

$\Large\boxed{\bf{Guide}}$

$\Large\textrm{Trigonometric Ratios}$

"The ratio of two sides of a right triangle is called a trigonometric ratio."

Let's solve this later to learn some information.

$\cdots\longrightarrow\boxed{\sin\theta=\dfrac{\rm{Opp.}}{\textrm{Hyp.}}}$

$\cdots\longrightarrow\boxed{\cos\theta=\dfrac{\rm{Adj.}}{\textrm{Hyp.}}}$

$\cdots\longrightarrow\boxed{\tan\theta=\dfrac{\rm{Opp.}}{\textrm{Adj.}}}$

(Usually, we memorize it by calling SOH-CAH-TOA.)

$\Large\boxed{\bf{Solution}}$

$\large\textrm{Components of \triangle\rm{PQR} : -}$

$\begin{aligned}&\textrm{ \overline{\rm{QR}} : Adjacent}\\\\&\textrm{ \overline{\rm{PQ}} : Opposite}\\\\&\textrm{ \overline{\rm{RP}} : Hypotenuse}\end{aligned}$

$\Large\boxed{\bf{Answer}}$

$\sin\theta=\dfrac{\overline{\rm{PQ}}}{\overline{\rm{RP}}}\dots \rm{(I)}$

$\cos\theta=\dfrac{\overline{\rm{QR}}}{\overline{\rm{RP}}}\rm{\dots(II)}$

$\textrm{using Pythagoras theorem}$

$\overline{\rm{PQ}}^{2}+\overline{\rm{QR}}^{2}=\overline{\rm{PR}}^{2}$

$\textrm{dividing each term by \overline{\rm{PR}}^{2} }$

$\dfrac{\overline{\rm{PQ}}^{2}}{\overline{\rm{PR}}^{2}}+\dfrac{\overline{\rm{QR}}^{2}}{\overline{\rm{PR}}^{2}}=\dfrac{\overline{\rm{PR}}^{2}}{\overline{\rm{PR}}^{2}}$

$\left(\dfrac{\overline{\rm{PQ}}}{\overline{\rm{PR}}}\right)^{2}+\left(\dfrac{\overline{\rm{QR}}}{\overline{\rm{PR}}}\right)^{2}=1$

$\textrm{ \sin^{2}\theta+\cos^{2}\theta=1 , from I and II}$

I hope this helped.