Question

From the given figure, evaluate :
(1) sin2 B + cos2 B (1) cos B
(in) tan C
(iv) sin B cos C + cos B sin C​

Answer 1

Given :-

• In Right angle ∆ ABC , ∠A = 90°
• In which AB = 8. BC = 17 , AC = ?

To Find :-

• (i) Sin²B + Cos²B = ?
• (ii) CosB = ? (iii) TanC = ?
• (iv) SinB × CosC + CosB × SinC = ?

Solution :-

To evaluate the above expression , at first we have to find the the length of AC. By applying Pythagoras theorem Then evaluate the above expression. from given question , we have to assume AB is perpendicular , BC is hypotenuse and

AC is base of the given triangle.

As per Pythagoras theorem :-

⇒ BC² = AB² + AC²

⇒ 17² = 8² + AC²

⇒ 289 = 64 + AC²

⇒ AC² = 289 - 64

⇒ AC = √225

⇒ AC = 15

Now evaluate the expression here :-

⇒ Sin²B + Cos²B

⇒ (AC/BC)² + (AB/BC)²

⇒ (15/17)² + (8/17)²

⇒ 225/289 + 64/289

⇒ (225 + 64)/289

⇒ 289/289 = 1

⇒ CosB = (AB/BC)

⇒ CosB = 8/17

⇒ TanC = (AB/AC)

⇒ TanC = 8/15

⇒ SinB × CosC + CosB × SinC

⇒ (AC/BC) × (AC/BC) + (AB/BC) × (AB/BC)

⇒ 15/17 × 15/17 + 8/17 × 8/17

⇒ 225/289 + 64/289

⇒ (225 + 64)/289

⇒ 289/289 = 1

Some note related to this question :-

• As per theta C :-

⇒ Theta C = adjacent side is AC

⇒ Theta C = opposite side is AB

• As per theta B :-

⇒ Theta B = adjacent side is AB

⇒ Theta B = opposite side is AC

• As per theta A :-

⇒ Theta A = adjacent side is AB or AC

⇒ Theta A = opposite side is AB or AC

Answer 2

Given :-

A triangle ABC

Distance of AB = 8

Distance of BC = 17

To Find :-

(i) sin² B + cos² B

(ii)cos B

(iii) tan C

(iv) sin B cos C + cos B + sin C

Solution :-

At first we need to know some formula

${\large{\boxed{\bf{\dag sin(\theta)=\dfrac{opp}{hyp}}}}}$

${\large{\boxed{\bf{\dag cos(\theta)=\dfrac{adj}{hyp}}}}}$

${\large{\boxed{\bf{\dag tan(\theta)=\dfrac{opp}{adj}}}}}$

By Pythagoras theorem

${\large{\boxed{\bf{\dag (BC)^2=(AB)^2+(AC)^2}}}}$

⇒ (17)² = (8)² + (AC)²

⇒ (17 × 17) = (8 × 8) + (AC)²

⇒ 289 = 64 + AC²

⇒ AC² = 289 - 64

⇒ AC² = 225

⇒ AC = √225

⇒ AC = 15 cm

(i) sin² B + cos²B

sin θ = opp/hyp

sin θ = 15/17

cos θ = adj/hyp

cos θ = 8/17

(15/17)² + (8/17)²

(225/289) + (64/289)

225 + 64/289

289/289

1

(ii) cos B

cos θ = adj/hyp

cos θ = 8/17

(iii) tan C

tan θ = opp/adj

tan θ = 8/15

(iv) sin B cos C + cos B sin C

(15/17) × (15/17) + (8/17) × (8/17)

225/289 + 64/289

289/289

1