Question

How can you explain: sin²∅+cos²∅=c²/b²+a²/b²=c²+a²/b²= b²/b²=1

Answer 1

Step-by-step explanation:

Given that,

\rm \:\angle SPR = 25\degree∠SPR=25°

\rm \:\angle PRS = 60\degree∠PRS=60°

In \triangle△ PRS,

We know, sum of all interior angles of a triangle is supplementary.

So, using this

\begin{gathered}\rm \: \angle PRS + \angle SPR + \angle RSP = 180\degree \\ \end{gathered}

∠PRS+∠SPR+∠RSP=180°

\begin{gathered}\rm \: 60\degree + 25\degree + \angle RSP = 180\degree \\ \end{gathered}

60°+25°+∠RSP=180°

\begin{gathered}\rm \: 85\degree + \angle RSP = 180\degree \\ \end{gathered}

85°+∠RSP=180°

\begin{gathered}\rm \: \angle RSP = 180\degree - 85\degree \\ \end{gathered}

∠RSP=180°−85°

\begin{gathered}\rm \: \bf\implies \:\angle RSP = 95\degree \\ \end{gathered}

⟹∠RSP=95°

Now, PQRS is a cyclic quadrilateral.

We know, sum of the opposite angles of a cyclic quadrilateral is supplementary.

So, using this, we get

\begin{gathered}\rm \: \angle RSP + \angle RQP = 180\degree \\ \end{gathered}

∠RSP+∠RQP=180°

\begin{gathered}\rm \: 95\degree + \angle RQP = 180\degree \\ \end{gathered}

95°+∠RQP=180°

\begin{gathered}\rm \: \angle RQP = 180\degree - 95\degree \\ \end{gathered}

∠RQP=180°−95°

\begin{gathered}\rm\implies \:\boxed{ \rm{ \:\bf \: \angle RQP = 85\degree \: \: }} \\ \end{gathered}

⟹

∠RQP=85°