Question

The
∆RPQ is right angled at Q
PQ=5cm. and RQ=10cm, find
& SinP,cosP, CosR, tanR​

Answer 1

Answer:

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Answer 2

⠀⠀⠀⠀⠀$\huge\underline{ \mathrm{ \red{QueS{\pink{tiOn}}}}}$

The

∆RPQ is right angled at Q

PQ=5cm. and RQ=10cm, find

& SinP,cosP, CosR, ttanR

_______________________________________________

⠀⠀⠀⠀⠀⠀$\huge{ \underline{ \purple{ \bold{ \underline{ \mathrm{ExPlanA{\green{TiOn }}}}}}}}$

$\large\underline{ \underline{ \blue{ \bold {Given}}}} = >$

⠀⠀⠀⠀※ A right angled triangle⠀RPQ.

⠀⠀⠀⠀※ PQ = 5cm

⠀⠀⠀⠀※ PQ = 5cm⠀⠀⠀⠀

______________

$\large\underline{ \underline{ \blue{ \bold{To\:Find}}}} = >$

⠀⠀⠀⠀⠀※ sin P

⠀⠀⠀⠀⠀※ cos P

⠀⠀⠀⠀⠀※ cos R

⠀⠀⠀⠀⠀※ tan R

______________

$\large\underline{ \underline{ \blue{ \bold{solution}}}} = >$

⠀⠀⠀⠀⠀By Pythagoras theorem

⠀⠀⠀⠀⠀

$\bold {RP {}^{2} =RQ {}^{2} +PQ {}^{2} }$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀$\bold{RP {}^{2} = 10 {}^{2} + 5 {}^{2} } \\ \\ \bold{RP = \sqrt{125} } \\ \\ \bold{RP = 5 \sqrt{5}cm }$

$\bold{sin \: P= \frac{ RQ}{RP}} \\ \\ = > \: \: \: \: \: \: \frac{10}{5 \sqrt{5} } \\ \\ = > \: \: \: \: \: \: \frac{2}{ \sqrt{5} }$

⠀⠀⠀⠀⠀$\bold{cos \:P= \frac{ QP }{RP} } \\ \\ = > \: \: \: \: \frac{5}{5 \sqrt{5} } \\ \\ = > \: \: \: \: \frac{1}{ \sqrt{5} }$

$\bold{Cos \: R= \frac{RQ}{RP} } \\ \\ = > \: \: \: \: \frac{10}{5 \sqrt{5} } \\ \\ = > \: \: \: \: \frac{2}{ \sqrt{5} }$

⠀⠀⠀⠀⠀$\bf \: Tan \: R= \frac{ QP }{RP} \\ \\ = > \bf\: \: \: \: \frac{5}{10} \\ \\ = > \: \: \: \: \frac{1}{2}$

______________________________

hops this may help you

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