Question

in a triangle PQR right angled at Q, PR+QR= 25cm and PQ=5cm. Determine the value of sin p, cos p, tan p​

Answer 1

Required Answer:

Given:

• PQR is a right-angled triangle, Angle Q = 90°.

• PR + QR = 25cm

• PQ = 5cm

To find:

• Value of sin P, cos P , tan P.

Calculation:

Here,

• PQ ( Base ) = 5cm
• PR ( Hypotenuse )
• PQ ( Perpendicular )

Let us assume PQ as x. Thus,

• PR becomes 25 - x

Now, by applying pythagoras property:

$\longrightarrow$ H² = B² + P²

$\longrightarrow$ ( 25 - x )² = (5)² + (x)²

$\longrightarrow$ 625 + x² - 50x = 25 + x²

$\longrightarrow$ 625 - 25 - 50x = x² - x²

$\longrightarrow$ 600 - 50x = 0

$\longrightarrow$ -50x = 0 - 600

$\longrightarrow$ -50x = -600

$\longrightarrow$ x = -600/-50

$\longrightarrow$ x = 12 cm

Now,

• PQ (x)= 12cm                 [Perpendicular]
• PR (25-x) → 25 - 12 → 13 cm [Hypo.]
• Base = 5cm

Also, we know that,

● sin $\theta$ = Perpendicular/Hypotenuse

$\longrightarrow$ sin P = 12/13

____________

● cos $\theta$ = Base/Hypotenuse

$\longrightarrow$ cos P = 5/13

____________

● tan $\theta$ = Perpendicular/Base

$\longrightarrow$ tan P = 12/5

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Extra info:

● sin $\theta$ = Perpendicular/Hypotenuse

● cosec $\theta$ = Hypotenuse/Perpendicular

● cos $\theta$ = Base/Hypotenuse

● sec $\theta$ = Hypotenuse/Base

● tan $\theta$ = Perpendicular/Base

● cot $\theta$ = Base/Perpendicular

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