Question

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In A OPQ, right-angled at P,
OP = 7 cm and OQ - PQ = 1 cm (see Fig. 8.12).
Determine the values of sin Q and cos Q.​

Answer 1

Answer:

Step-by-step explanation:

$\large{\mathtt{OQ-PQ=1 \: cm}}$

$\large{\mathtt{OQ=1+PQ}}$

Using Pythagoras theorem,

$\large{\mathtt{(a)^2+(b)^2=(c)^2}}$

$\large{\mathtt{(OP)^2+(PQ)^2=(OQ)^2}}$

$\large{\mathtt{(7)^2+(PQ)^2=(1+PQ)^2}}$

$\large{\mathtt{49+(PQ)^2=(PQ)^2+1+2PQ}}$

$\large{\mathtt{49+(PQ)^2-(PQ)^2-1-2PQ=0}}$

$\large{\mathtt{2PQ=48}}$

$\large{\mathtt{PQ=24}}$

$\large{\mathtt{OQ=1+PQ}}$

$\large{\mathtt{OQ=1+24}}$

$\large{\mathtt{OQ=25}}$

$\huge{\mathtt{sin(Q)=\frac{7}{25}}}$

$\huge{\mathtt{cos(Q)=\frac{24}{25}}}$

Answer 2

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In Δ OPQ, we have

OQ^2=OP^2+PQ^2

⇒(PQ+1)^2=OP^2+PQ^2 [∵OQ−PQ=1⇒OQ=1+PQ]

⇒PQ^2+2PQ+1=OP^2+PQ^2

⇒2PQ+1=49

⇒PQ=24cm

∴OQ−PQ=1cm

⇒OQ=(PQ+1)cm=25cm

Now, sinQ=OQ/OP=25/7

and, cosQ= OQ/PQ=25/24