Question

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

Answer 1

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

A triangle PQR, right angled at Q, PR + QR = 25 cm, and PQ = 5cm.

$\purple{\large\underline{\sf{To\:Find - }}}$

The values of sinP and cosP

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

Given that,

A triangle PQR, right angled at Q, PR + QR = 25 cm, and PQ = 5cm.

Let assume that, QR = x cm

So, PR = 25 - x cm

Now, In right triangle PQR, using Pythagoras Theorem,

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

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

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

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

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

$\rm :\longmapsto\: 5x = 60$

$\bf\implies \:x = 12$

Hence, we have now

In triangle PQR

• PQ = 5 cm

• QR = x = 12 cm

• PR = 25 - x = 25 - 12 = 13 cm

So,

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

and

$\rm :\longmapsto\:cosP = \dfrac{QP}{PR} = \dfrac{5}{13}$

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