If AB 7 cm and BC- AC 1 cm, find :

(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

BC² = AB² + AC²

Substituting the value of BC

(1 + AC)² = AB² + AC²

⇒ 1 + AC² + 2AC = 7² + AC²

1 + AC² + 2AC = 49 AC²

⇒ 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

PR² = PQ² + QR²

⇒ (50 – QR)² = (40)² + QR²

2500 + QR² – 100QR = 1600 + QR²

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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