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

Hello, Buddy!!

➸ ɢɪᴠᴇɴ:-

• ABC is a Right Angle Triangle, ∠BAC = 90°.
• Length of AB = 8 units.
• Length of BC = 17 units.

➸ ᴛᴏ ꜰɪɴᴅ:-

• sin²B+cos²B
• cosB
• tanC
• sinBcosC+cosBsinC

➸ ʀᴇQᴜɪʀᴇᴅ ꜱᴏʟᴜᴛɪᴏɴ:-

As, it's a Right Angle Triangle

By Pythagoras Theorem

→ (side)²+(side)² = (Hypothenuse)²

→ (AB)²+(AC)² = (BC)²

→ (8)²+(AC)² = (17)²

→ (AC)² = 289-64

→ AC = √(225)

→ AC = √(25)²

→ AC = 15 units.

• Length of AC = 15 units.

→ sin²B+cos²B

→ (AC/BC)²+(AB/BC)²

→ (15/17)²+(8/17)²

→ (225/289)+(64/289)

→ (225+64)/289

→ 289/289

→ 1

• sin²B+cos²B $\mapsto\;\Large{\red{\bold{1}}}$

→ cosB

→ (AB/BC)

→ 8/17

• cosB $\mapsto\;\Large{\green{\bold{\frac{8}{17}}}}$

→ tanC

→ (AB/AC)

→ 8/15

• tanC $\mapsto\;\Large{\purple{\bold{\frac{8}{15}}}}$

→ sinBcosC+cosBsinC

→ (AC/BC)×(AC/BC)+(AB/BC)×(AB/BC)

→ (15/17)×(15/17)+(8/17)×(8/17)

→ (15/17)²+(8/17)²

→ (225/289)+(64/289)

→ (225+64)/289

→ 289/289

• sinBcosB+cosBsinC $\mapsto\;\Large{\blue{\bold{1}}}$

☛ ƒσɾɱµℓαร µรε∂:-

• Sin theta = Opposite/Hypothenuse
• Cos theta = Adjacent/Hypothenuse
• Tan theta = Opposite/Adjacent

$\boxed{\tt{@MrSovereign♡}}$

Hope This Helps!!

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 + cos² B

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