Question

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Answer 1

Answer:

answer of (c)

Step-by-step explanation:

AB=5cm as AC=5cm

THEREFORE,

AD=4cm

BD=AB^2-AD^2=3cm

BD=3cm

THEREFORE,

SEC O = AB/AD = 5/4 cm

COS(90°-O) = SIN O = BD/AD = 3/4 cm

SO AT LAST,

SEC O + COS(90°-O)

= 5/4 + 3/4

= 8/4

= 2

Answer 2

Question :

In the adjoining figure, ABC is an isosceles triangle in which AB = AC. Find the values of the following :

• tan θ

• sin (90° - θ)

• sec θ + cos (90° - θ)

Solution :

According to the question :

AB = AC

AC = 5 cm [given]

So putting it's value we get

AB = 5 cm

So, AB is 5 cm

Now, we need to apply Pythagoras Theorem

In rt∠d ∆ADC, ∠D is 90°

(Hypotentuse)² = (Perpendicular)² + (Base)²

(AC)² = (AD)² + (CD)²

Here,

AC = 5 cm

CD = 3 cm

AD = ?

Putting their respective values we get

(AD)² = (5)² - (3)²

(AD)² = 25 - 9

(AD)² = 16

AD = √16

AD = 4 cm

Now, to get the the value of base in ∆ABC using Pythagoras Theorem

In rt∠d ∆ABC, ∠D is 90°

(Hypotentuse)² = (Perpendicular)² + (Base)²

(AB)² = (AD)² + (BC)²

Here,

AB = 5 cm

AD = 4 cm

BC = ?

Putting all the values we get

(5)² = (4)² + (BC)²

(BC)² = 25 - 16

(BC)² = 9

BC = √9

BC = 3

Now according to the given information we need to find :

1) tan θ

We know that,

• tan θ = perpendicular/adjacent

• tan θ = 3/4

2) sin (90° - θ)

We know that,

sin (90° - θ) = cosec θ

So formula for cos θ is

• cos θ = hypotentuse/base

• cosec θ = 5/4

3) sec θ + cos (90° - θ)

We know that,

cos(90° - θ) = sec θ

Formula for sec θ

• sec θ = Hypotentuse/Base

• sec θ = 5/4

Now,

sec θ + sec θ

2 sec θ

2 × 5/4

5/2

Final Answers :

1) tan θ = 3/4

2) sin (90° - θ) = 5/4

3) sec θ + cos (90° - θ) = 5/2

Learn More :

Basic :

sin θ = P/H

cos θ = B/H

tan θ = P/B

cot θ = B/P

sec θ = H/B

cosec θ = H/P

Here,

P refers Perpendicular or Height

B refers Base

H refers Hypotentuse

Square Relations :

sin² θ + cos² θ = 1

sec² θ – tan² θ = 1

cosec² θ – cot² θ= 1

Quotient Relations :

sin θ× cosec θ = 1

cos θ × sec θ = 1

tan θ × cot θ = 1

Trigonometric value of standard angles :

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin θ & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos θ & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan θ & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec θ & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec θ & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot θ & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$

Trigonometrical ratios of complementary angles :

sin (90° - θ) = cosec θ

cos (90° - θ) = sec θ

tan (90° - θ) = cot θ

cot (90° - θ) = tan θ

sec (90° - θ) = cos θ

cosec (90° - θ) = sin θ