Question

$\huge \fbox \red{❥ Question}$


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


(REQUIRED QUALITY ANSWER)​

Answer 1

ANSWER:

➠PQ=5cm

➠PR+QR=25cm

➠PR=25cm-QR.....(1)

In ∆PQR applying Pythagoras theorem

➠(PR)²=PQ²+QR²

Put value of (1) equation in this statement

➠(25-QR)²=5²+QR²

➠625+QR²-50QR=25+QR²

➠-50QR+625-25=QR²-QR²

➠600=50QR

$\sf➠ \frac{600}{50 } = QR$

➠12cm=QR

➠PR=25-12=13cm

$\sf★ \: Sin \: P= \frac{QR}{PR}=\frac{12}{13}$

$\sf ★Tan \: P=\frac{QR}{PR}=\frac{12}{5}$

$\sf ★Cos \: P=\frac{PQ}{PR}=\frac{5}{13}$

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf 5cm}\put(2.8,.3){\large\bf }\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf P}\put(.8,.3){\large\bf Q}\put(5.8,.3){\large\bf R}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf \Theta }\end{picture}$

Kindly open web to see diagram.

Answer 2

Answer:

ur answer is here hope it helps

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