**(a) In ..ABC, A = 90°. If AB 7 cm and BC- AC 1 cm, find :**

**(1) sin C**

**(i) tan B**

**(b) In …PQR,Q = 90°. If PQ = 40 cm and PR + QR = 50 cm, find :**

**(1) sin P**

**(ii) cos P**

**(ti) tan R**

**Answer :**

**(a)** In right ∆ABC

∠A = 90°

AB = 7 cm

BC – AC = 1 cm

⇒ BC = 1 + AC

BC^{2} = AB^{2} + AC^{2}

Substituting the value of BC

(1 + AC)^{2} = AB^{2} + AC^{2}

⇒ 1 + AC^{2} + 2AC = 7^{2} + AC^{2}

1 + AC^{2} + 2AC = 49 AC^{2}

⇒ 2AC = 49 – 1 – 48

AC = 48/2 = 24 cm

Here,

BC = 1 + AC

BC = 1 + 24 = 25 cm

**(i)** sin C = AB/BC = 7/25

**(ii)** tan B = AC/AB = 24/7

**(b)** In right ∆PQR

∠Q = 90°

PQ = 40 cm

⇒ PQ + QR = 50 cm

We can write it as

PQ = 50 – QR

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

⇒ (50 – QR)^{2} = (40)^{2} + QR^{2}

2500 + QR^{2} – 100QR = 1600 + QR^{2}

So we get

2500 – 1600 = 100QR

⇒ 100QR = 900

QR = 900/100 = 9

We get

PR = 50 – 9 = 41

**(i)** sin P = QR/PR = 9/41

**(ii)** cos P = PQ/PR = 40/41

**(iii) **tan R = PQ/QR = 40/9

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