Question

10. In A PQR, right-angled at Q. PR + QR = 25 cm and PQ = 5 cm. Determine the values of
sin P. cos P and tan P.
+
11 Stale whether the following are true​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, In triangle PQR, right angled at Q,

• PR + QR = 25 cm

• PQ = 5 cm

Let assume that,

QR = x cm

So, PR = 25 - x cm

So, In right triangle PQR

By, using Pythagoras Theorem, we have

$\rm :\longmapsto\: {PR}^{2} = {PQ}^{2} + {QR}^{2}$

$\rm :\longmapsto\: {(25 - x)}^{2} = {x}^{2} + {5}^{2}$

$\rm :\longmapsto\:625 + {x}^{2} - 50x = {x}^{2} + 25$

$\rm :\longmapsto\:625 - 50x = 25$

$\rm :\longmapsto\: - 50x = 25 - 625$

$\rm :\longmapsto\: - 50x = - 600$

$\bf\implies \:x = 12$

Thus, we have

$\rm :\longmapsto\:PR = 13 \: cm$

$\rm :\longmapsto\:QR = 12 \: cm$

$\rm :\longmapsto\:PQ = 5\: cm$

So,

$\rm\implies \:sinP = \dfrac{QR}{PR} = \dfrac{12}{13}$

$\rm\implies \:cosP = \dfrac{PQ}{PR} = \dfrac{5}{13}$

$\rm\implies \:tanP = \dfrac{QR}{PQ} = \dfrac{12}{5}$

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1