Question

in triangle 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 :

$\sf\underline{\red{\:\:\: Given:-\:\:\:}}$

• PR + QR = 25 cm
• PQ = 5 cm

$\sf\underline{\red{\:\:\: Need\:To\: Find:-\:\:\:}}$

• Values of Sin P, Cos P & tan P

$\bf{\underline{\underline \blue{Explanation:-}}}$

$\star$ Reference of image is in diagram.

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{P}}\put(7.7,1){\large\sf{Q}}\put(10.6,1){\large\sf{R}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.1,2){\sf{\large{5 cm}}}\put(9,0.7){\sf{\large{12}}}\put(9.4,1.9){\sf{\large{13 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

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Let us consider that QR be x & PQ = 5cm. (Given)

$\dashrightarrow\sf PR + QR = 25 cm \\ \\ \dashrightarrow\sf PR = 25 - QR \:\:\:\: \Big[ QR = x \Big] \\ \\ \dashrightarrow\sf PR = 25 - x$

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Now, In ∆ PQR

$\sf\underline{\red{\:\:\: By \: Using \: Pythagoras \: Theorem \: :-\:\:\:}}$

$\sf{Here}\begin{cases}\sf{PR \; (25 - x) \: is \: Hypotenuse} \\ \sf{PQ \:(5 \: cm) \: is \; Height} \\ \sf{QR \: (x \: cm) \: is \: Base} \end{cases}$

$\dashrightarrow\sf \Big(Hypotenuse \Big)^2 = {\Big(Height \Big)}^2+ {\Big(Base \Big) }^2\\ \\ \dashrightarrow\sf PR^2 = PQ^2 + QR^2 \\ \\ \dashrightarrow\sf \Big(25 - x \Big) = 5^2 + x^2 \\ \\ \dashrightarrow\sf (25)^2 + x^2 - 2 \times 25 \times x = 25 + x^2 \\ \\ \dashrightarrow\sf 625 + x^2 - 50 x = 25 + x^2 \\ \\ \dashrightarrow\sf 625 + \cancel{x^2} - 50 x - 25 - \cancel{ x^2 } = 0 \\ \\ \dashrightarrow\sf - 50 x + 600 = 0 \\ \\ \dashrightarrow\sf - 50x = - 600 \\ \\ \dashrightarrow\sf x = \cancel\dfrac{-600}{-50} \\ \\ \dashrightarrow\sf\red{x = 12}$

We get, the value of x.

And, Substituting the value of x.

$\dashrightarrow\sf PR = 25 - x \\ \\ \dashrightarrow\sf PR = 25 - 12 \\ \\ \dashrightarrow\sf\red{PR = 13 \: cm}$

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$\dashrightarrow\sf Sin P = \dfrac{QR}{PR} \;\;\;\; \bigg(QR = 12 \: cm \: \& PR = 13 \: cm \bigg)\\ \\ \dashrightarrow\sf \dfrac{12}{13}$

$\dashrightarrow\sf Cos P = \dfrac{PQ}{PR} \\ \\ \dashrightarrow\sf \dfrac{5}{13}$

$\dashrightarrow\sf tan P = \dfrac{QR}{PQ} \\ \\ \dashrightarrow\sf \dfrac{12}{5}$

Answer 2

Answer:

Given :-

PR + QR = 25 and PQ = 5

Let PR be x.

then,

QR = 25 - x

Using Pythagoras theorem :

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

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

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

50x = 650

x = 13

Thus,

PR = 13 cm

and, QR = (25 - 13) cm = 12 cm

Now,

sin P = QR/PR = 12/13

sin P = QR/PR = 12/13cos P = PQ/PR = 5/13

sin P = QR/PR = 12/13cos P = PQ/PR = 5/13tan P = QR/PQ = 12/5