**(a) In the figure (1) given below, ∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm. Find**

(i) sin C

(ii) tan B

(iii) tan C – cot B.

**(b) In the figure (2) given below, ∆ABC is right-angled at B. Given that ∠ACB = θ, side AB = 2 units and side BC = 1 unit, find the value of sin ^{2} θ + tan^{2} θ.**

**(c) In the figure (3) given below, AD is perpendicular to BC, BD = 15 cm, sin B = 4/5 and tan C = 1.**

(i) Calculate the lengths of AD, AB, DC and AC.

(ii) Show that tan^{2} B – 1/cos^{2} B = – 1.

**Answer :**

##### (a) ∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm

Construct AD perpendicular to BC

D is the mid point of BC

So BD = CD

BD = CD = 6/2 = 3 cm

In right angled ∆ABD

AB^{2} = AD^{2} + BD^{2}

AD^{2} = AB^{2} – BD^{2}

AD^{2} = 5^{2} – 3^{2}

AD^{2} = 25 – 9 = 16

AD^{2} = 4^{2}

AD = 4 cm

**(i)** In right angled ∆ACD

sin C = perpendicular/hypotenuse

sin C = AD/AC = 4/5

**(ii)** In right angled ∆ABD

tan B = perpendicular/base

tan B = AD/BD = 4/3

**(iii)** In right angled ∆ACD

tan C = perpendicular/base

⇒ tan C = AD/CD = 4/3

In right angled ∆ABD

cot B = base/perpendicular

⇒ cot B = BD/AD = ¾

tan C – cot B = 4/3 – ¾

Taking LCM,

tan C – cot B = (16 – 9)/12 = 7/12

**(b)** It is given that

∆ABC is right-angled at B

AB = 2 units and BC = 1 unit

In right angled ∆ABC

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

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

⇒ AC^{2} = 4 + 1 = 5

AC^{2} = 5

⇒ AC = √5 units

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

⇒ sin θ = AB/AC = 2/√5

In right angled ∆ABC

tan θ = perpendicular/base

⇒ tan θ = AB/BC = 2/1

sin^{2} θ + tan^{2} θ = (2/√5)^{2} + (2/1)^{2}

= 4/5 + 4/1

Taking LCM

= (4 + 20)/5

= 24/5

= 4 4/5

**(c) (i)** In ∆ABC

AD is perpendicular to BC

BD = 15 cm

sin B = 4/5

tan C = 1

In ∆ABD

sin B = perpendicular/hypotenuse

⇒ sin B = AD/AB = 4/5

Consider AD = 4x and AB = 5x

In right angled ∆ABD,

AB^{2} = AD^{2} + BD^{2}

BD^{2} = AB^{2} – AD^{2}

(15)^{2} = (5x)^{2} – (4x)^{2}

⇒ 225 = 25x^{2} – 16x^{2}

225 = 9x^{2}

⇒ x^{2} = 225/9 = 25

x = √25 = 5

AD = 4 × 5 = 20

AB = 5 × 5 = 25

In right angled ∆ACD

tan C = perpendicular/base

tan C = AD/CD = 1/1

Consider AD = X then CD = x

In right angled ∆ADC

AC^{2} = AD^{2} + CD^{2}

AC^{2} = x^{2} + x^{2} **…(1)**

So the equation becomes

AC^{2} = 20^{2} + 20^{2}

⇒ AC^{2} = 400 + 400 = 800

AC = √800 = 20√2

Length of AD = 20 cm

Length of AB = 25 cm

Length of DC = 20 cm

Length of AC = 20√2 cm

**(ii)** In right angled ∆ABD

tan B = perpendicular/base

tan B = AD/BD

Substituting the values

tan B = 20/15 = 4/3

In right angled ∆ABD,

cos B = base/hypotenuse

cos B = BD/AB

Substituting the values

cos B = 15/25 = 3/5

Here,

LHS = tan^{2} B – 1/cos^{2} B

= (4/3)^{2} – 1/(3/5)^{2}

= (4)^{2}/(3)^{2} – (5)^{2}/(3)^{2}

= 16/9 – 25/9

So we get

= (16 – 25)/9

= -9/9

= – 1

= RHS

Hence, proved.

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