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# In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P

**Solution:**

We will use the trigonometric ratios to solve the question.

Using the Pythagoras theorem, we can find the length of all three sides. Then, we will find the required trigonometric ratios.

Given ∆PQR is right-angled at Q.

PQ = 5 cm

PR + QR = 25 cm

Let PR = x cm

Therefore,

QR = 25 cm - PR

= (25 - x) cm

By applying Pythagoras theorem in ∆ PQR, we obtain.

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

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

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

50x = 650

x = 650/50 = 13

Therefore, PR = 13 cm

QR = (25 - 13) cm = 12 cm

By substituting the values obtained above in the trigonometric ratios below we get,

sin P = side opposite to angle P / hypotenuse = QR/PR = 12/13

cos P = side adjacent to angle P / hypotenuse = PQ/PR = 5/13

tan P = side opposite to angle P / side adjacent to angle P = QR/PQ = 12/5

**☛ Check: **Class 10 Maths NCERT Solutions Chapter 8

**Video Solution:**

## In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Maths NCERT Solutions Class 10 Chapter 8 Exercise 8.1 Question 10

**Summary:**

In ΔPQR, right-angled at Q, if PR + QR = 25 cm and PQ = 5 cm, then the values of sin P, cos P, and tan P are 12/13, 5/13, and 12/5 respectively.

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