Question

Hello all can anybody explain me the chapter trigonometry plz done give unnecessary answers I u will give unnecessary answer it would be bad plz don't do that if ur doing for points I will make other questions. plz explain total chapter.​

Answer 1

★ Answer :

Firstly, for starting Chapter trignometry you should know about tge basics of this chapter.

You have to learn the following thing firstly,

theta (θ) means angle. Theta can be any angle. According to class 10 value of theta can be 0°, 30°, 45°, 60° and 90°.

→ There are 6 basic terms in chapter 8 & 9 respectively.

➳ Sine θ

➳ Cosine θ

➳ tangent θ

➳ Cosecent θ

➳ Secent θ

➳ cotangent θ

We write Sine θ as Sin θ, Cosine θ as Cos θ, tangent θ as tan θ, Cosecant θ as cosec θ, Secent θ as Sec θ and cotangent θ as cot θ repectively.

In first exercise of the chapter. We will be given value of some angle such as sin θ and we have to find the value of 5 others.

$\rule{200}{2}$

Firstly, we shoulf know that,

Hypotenuse : Hypotenuse is the side which is touvhed to the angle which is given expect 90°. (Longest side).

Base : In the line where both angles will be touched are known as base.

Perpendicular : On the line where there os only 90° angle that will be perpendicular.

→ Value of Sin θ = Perpendicular/Hypotenuse

→ Value of cos θ = Base/Hypotenuse

→ Value of tan θ = Perpendicular/Base

Value of cosec θ will be reciprocol of value of sin θ such as sin θ = 1/2. Then cosec θ = 2. Similarly, sec θ will be reciprocol of value of cos θ and cot θ will be reciprocol of value of tan θ.

So,

→ Value of cosec θ = 1/sin θ = 1/Perpendicular/Hypotenuse = Hypotenuse/Perpendicular

→ Value of sec θ = 1/cos θ = 1/Base/Hypotenuse = Hypotenuse/Base

→ Value of cot θ = 1/tan θ = 1/Perpendicular/Base = Base/Perpendicular.

$\rule{200}{2}$

As, I told you earlier that value of cosec θ, sec θ and cot θ are rciprocol.

$\rule{200}{2}$

Now, There are sum indentities in class 10 chapter Trigonometry which were given below ⤵⤵⤵

(i) Sin (90° - θ) = Cos θ

(ii) Cos (90° - θ) = Sin θ

(iii) tan (90° - θ) = Cot θ

(iv) Cot (90° - θ) = tan θ

(v) Sec (90° - θ) = Cosec θ

(vi) Cosec (90° - θ) = Sec θ

(vii) Sin²θ + Cos²θ = 1

(viii) 1 + tan²θ = sec²θ

(ix) 1 + Cot²θ = Cosec²θ

Answer 2

For explaining you iam giving a question and solving that :

$\bf{\underline{\green{Question: - }}}$

in a right angled triangle ∆ ABC

A = 90° and $\frac{sin \: b}{cos \: b}$= 2k find the value of k

$\bf{\underline{\green{Answer: - }}}$

Diagram is in the image.

Sin B = $\frac{perpendicular}{hypotenuse} = \frac{ac}{bc}$

Cos C = $\frac{base}{hypotenuse} = \frac{ac}{bc}$

Now,

$\frac{ sin \: b }{cos \: c}$= 2k

$\frac{ac}{bc} \: \frac{bc}{ac} \: \frac{ac}{bc}$= 2k

1 = 2k k$= \frac{1}{2}$

so we get the answer $= \frac{1}{2}$

Hence verified !