Ex 8.1 Q10 - In ∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sinP, cosP and tanP.
Solution:-Given,
PQ = 5 cm
PR + QR = 25 cm
∴ PR = 25 -QR ----- (1)
➙ Applying Pythagoras theorem in ∆PQR,
(Hypotenuse)² = (Base)² + (Altitude)²
∴ PR² = QR² + PQ²
∴ (25 - QR)² = QR² + (5)² -- from Eqⁿ(1)
∴ 625 - 50QR + QR² = 25 + QR²
∴ 50QR = 625 - 25
∴ 50QR = 600
∴ QR =
600
/
50
∴ QR = 12 cm
➙ Putting QR = 12 in Eqⁿ(1)
∴ PR = 25 - 12
∴ PR = 13 cm
Now,
*sinP =
Opposite side of ∠P
/
Hypotenuse
=
QR
/
PR
=
12
/
13
∴ sinP =
12
/
13
*cosP =
Adjacent side of ∠P
/
Hypotenuse
=
PQ
/
PR
=
5
/
13
∴ cosP =
5
/
13
*tanP =
Opposite side of ∠P
/
Adjacent side of ∠P
=
QR
/
PR
=
12
/
5
∴ tanP =
12
/
5
Alternatively,
*tanP =
sinP
/
cosP
=
12/13
/
5/13
=
12
/
5
∴ tanP =
12
/
5
Hence,
The values of sinP, cosP and tanP are 12/13, 5/13 and 12/5 respectively.
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Try This..
✒ In triangle PQR, right-angled at Q, if PQ = 8 cm and the difference between PR and QR is 6 cm, calculate sin P, cos P, and tan P.
✒ A triangle XYZ is right-angled at Y. If XY = 7 cm and XZ – YZ = 11 cm, find the trigonometric ratios of angle X.
✒ In triangle ABC, right-angled at B, if AB = 9 cm and the sum of sides AC and BC is 30 cm, find sin A, cos A, and tan A.
❌ Common Mistakes
Misidentifying sides:▸ Always label opposite, adjacent, and hypotenuse correctly based on the reference angle.Incorrect substitution:▸ Plugging in wrong values in Pythagoras’ Theorem or forgetting to square the terms.Skipping verification:▸ Always double-check that your triangle satisfies the Pythagorean identity.
Queries Solved:-
Class 10 Ex 8.1
Ex 8.1 Q10 Class 10
Class 10 Ex 8.1 Q10
Class 10 Chap 8 Ex 8.1 Q10
Class 10 Introduction To Trigonometry
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