Question

In angle PQR, right-angled at Q, PR+QR=25cm and PQ=5cm. Determine the values of sin P, cos P and tan P.

Answer 1

Answer

The required values are

sinP = 12/13

cosP = 5/13

tanP = 12/5

Given

• ∆PQR is a right angled triangle
• PR + QR = 25cm and PQ = 5cm

To Find

The values of sinP , cosP and tanP

Formula to be used

• Pythagorean theorem : $\sf{perpendicular{}^{2} + base ^{2} = hypotenus^{2} }$
• $\sin \sf\theta = \sf \frac{perpendicular}{hypotenus}$
• $\cos \theta = \sf \frac{base}{hypotenus}$
• $\tan \theta = \sf \frac{ perpendicular}{base}$

Solution

Using the Pythagorean theorem we have :

$\sf{PQ ^{2} + QR {}^{2} =PR {}^{2} } \\ \implies \sf{PQ ^{2} =PR {}^{2} - QR {}^{2} } \\ \sf \implies {(5)}^{2} = (PR + QR)(PR - QR) \\ \sf \implies25 = 25(PR - QR) \\ \implies \sf PR - QR = 1 \: \: .........(1)$

Again , we are already given :

$\sf{PR + QR = 25}...........(2)$

Adding (1) and (2) we have :

$\sf{PR + QR + PR - QR = 25 + 1} \\ \implies \sf2PR = 26 \\ \implies \sf{PR = 13}$

Putting the value of PR in (2) we have :

$\sf \implies 13+ QR = 25 \\ \implies \sf{QR = 25 - 13} \\ \sf \implies QR =12$

Thus , the sides are :

PQ = 5cm

QR = 12cm

PR = 13cm

So Now required trigonometric functions are :

$\sf \sin(P) = \frac{QR}{PR} \\ \implies \sf\sin(P) = \frac{12}{13}$

$\sf\cos(P) = \frac{PQ}{PR} \\ \sf \implies \cos(P) = \frac{5}{13}$

$\sf\tan(P) = \frac{QR}{PQ} \\ \implies \sf{ \tan(P) = \frac{12}{5} }$

Answer 2

$\tt \huge{Answer}$

_______________________________

Question:

In angle PQR, right-angled at Q, PR+QR=25cm and PQ=5cm. Determine the values of sin P, cos P and tan P.

______________________________________

$\sf{Given}$

$\rm{PR+QR = 25}$

then

According to Question:

$\rm{ : \: ⟹let \: QR \: be \: x}$

$\rm{⟹PR+QR = 25}$

$\rm{⟹PR = 25 - QR}$

$\rm{⟹PR = 25 - x}$

________________________________

Now

$\tt{given \: right \: angled \: PQR}$

$\rm{⟹use \: the \: pythagoras \: theorem}$

$\bf{ \boxed{ \green{ \tt{( {hypotenuse})^{2} = ({height})^{2} + ({base})^{2} \: }}}}$

then

$\tt{ {PR}^{2} = {PQ}^{2} + {QR}^{2}}$

$\tt{⟹( {25 - x)}^{2} = {5}^{2} + {x}^{2}}$

$⟹\tt{( {25})^{2} + {x}^{2} - 2 \times 25 \times x = 25 + {x}^{2}}$

$\tt{⟹625 + {x}^{2} - 50x = 25 + {x}^{2}}$

$\tt{⟹625 + {x}^{2} - 50x - 25 - {x}^{2} = 0}$

$\tt{⟹ {x}^{2} + {x}^{2} - 50x + 625 - 25 = 0}$

$\tt{⟹ - 50 + 600 = 0}$

$\tt{⟹ - 50x = - 600}$

$\tt{⟹x = \frac{ - 600}{ - 50}}$

$\tt{⟹x = 12}$

so,

$\rm{QR = 12}$

$\rm{PR = 25 - x}$

$\rm{ = 25 - 12}$

$\rm{ = 13cm}$

• QR is 12cm

• PR is 13cm

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◕Now Determine the values of sin P, cos P and tan P.

❃Sinp

$\sf{sinp = \frac{side \: opp \: to \: angle \: p}{hyp}}$

$\sf{⟹ sinp = \frac{QR}{PR}}$

$\sf{⟹sinp = \frac{12}{13}}$

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◑cosp

$\sf{⟹cosp = \frac{side \: adj \: to \: angle \: p}{hyp}}$

$\sf{⟹cosp = \frac{PQ}{PR} }$

$\sf{⟹cosp = \frac{5}{13}}$

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◓tanp

$\sf{⟹tanp = \frac{side \: opp \: to \: p}{side \: adj \: to \: p}}$

$\sf{⟹tanp = \frac{QR}{PQ}}$

$\sf{⟹tanp = \frac{12}{15}}$

$\sf{⟹tanp = \frac{12}{5}}$

___________________________

so,

$⟹ \rm{tanp = \frac{sinp}{cosp}}$

$\rm{⟹tanp \frac{ \frac{12}{13} }{ \frac{5}{13}}}$

$\rm{⟹tanp = \frac{12}{5}}$

I hope it's help uh