Question

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◆ ɪɴ ∆ᴏᴘQ, ʀɪɢʜᴛ ᴀɴɢʟᴇᴅ ᴀᴛ ᴘ, ᴏᴘ = 7ᴄᴍ ᴀɴᴅ ᴏQ -ᴏᴘ = 1ᴄᴍ. ᴅᴇᴛᴇʀᴍɪɴᴇ ᴠᴀʟᴜᴇꜱ ᴏꜰ ᴀʟʟ ᴛʀɪɢᴏɴᴏᴍᴇᴛʀʏ ʀᴀᴛɪᴏꜱ ɪɴ ᴛᴇʀᴍꜱ ᴏꜰ ᴀɴɢʟᴇ Q.


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Answer 1

Answer:

In Δ OPQ, we have

$\sf \: OQ {}^{2} =OP {}^{2} +PQ {}^{2}$

$\sf⇒(PQ+1) {}^{2} =OP {}^{2} +PQ {}^{2} \bigg[∵OQ−PQ=1⇒OQ=1+PQ \bigg]$

$\sf \: ⇒PQ {}^{2} +2PQ+1=Op {}^{2} +PQ {}^{2}$

⇒2PQ+1=49

$\\ \\ \bold{PQ}=\frac{49-1}{2} \\ \\ \sf \sf \implies \bold{PQ}=\frac{48}{2} \\ \\ \sf \bold{⇒PQ=24cm}$

$\sf \: ∴OQ−PQ=1cm$

$\sf \: ⇒OQ=(PQ+1)cm=25cm$

$\sf \: Now, sinQ= \frac{OP}{ OQ}= \frac{7}{25} \\ \\ \sf \: and, cosQ= \frac{PQ}{ OQ}= \frac{24}{25}$

Hence, This is Answer.

Answer 2

Answer:

In Δ OPQ, we have

\sf \: OQ {}^{2} =OP {}^{2} +PQ {}^{2}OQ

2

=OP

2

+PQ

2

\sf⇒(PQ+1) {}^{2} =OP {}^{2} +PQ {}^{2} \bigg[∵OQ−PQ=1⇒OQ=1+PQ \bigg]⇒(PQ+1)

2

=OP

2

+PQ

2

[∵OQ−PQ=1⇒OQ=1+PQ]

\sf \: ⇒PQ {}^{2} +2PQ+1=Op {}^{2} +PQ {}^{2}⇒PQ

2

+2PQ+1=Op

2

+PQ

2

⇒2PQ+1=49

\begin{gathered}\\ \\ \bold{PQ}=\frac{49-1}{2} \\ \\ \sf \sf \implies \bold{PQ}=\frac{48}{2} \\ \\ \sf \bold{⇒PQ=24cm}\end{gathered}

PQ=

2

49−1

⟹PQ=

2

48

⇒PQ=24cm

\sf \: ∴OQ−PQ=1cm∴OQ−PQ=1cm

\sf \: ⇒OQ=(PQ+1)cm=25cm⇒OQ=(PQ+1)cm=25cm

\begin{gathered} \sf \: Now, sinQ= \frac{OP}{ OQ}= \frac{7}{25} \\ \\ \sf \: and, cosQ= \frac{PQ}{ OQ}= \frac{24}{25} \end{gathered}

Now,sinQ=

OQ

OP

=

25

7

and,cosQ=

OQ

PQ

=

25

24

Hence, This is Answer.