Question

Help! Maths, Class-10th
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Answer 1

Answer:

Q.1 Ans.sin theta is 5/13, cosec theta is 13/5,cos theta is 12/13, sec theta is 13/12,tan theta is 5/12 and cot theta is 12/5.

Step-by-step explanation:

Q.1. Ans

As sin is equal to perpendicular by hypotenus

therefore in the triangle the two sides i.e perpendicular and hypotenus is 5 and 13 rept.

Then we should find the other side i.e base of the triangle

by using Pythagorean theorem.Then the value comes out to be 12.After knowing the values of all the sides

we can find out the value of other trigonometric ratios

eg.tan=perpendicular / base

i.e tan=5/12.

Hope you can understand it.

If you can then you can solve the remaining questions as the methods used are same.

Answer 2

$\Large{\underline{\underline{\red{\tt{\purple{\leadsto } QuestionS:-}}}}}$

❒ Try Yourself :-

• 1) In a right angled ∆ sinθ = 5/13 . Find all the trignometric Ratios of the triangle .

• 2) Find (i) cotθ , when tanθ = 15/8 (ii) cscθ , when sinθ = √3/2 . (iii) secθ , when cosθ = 1/√2.

• 3) If sinθ = 3/5 and cosθ = 4/5 , then find tanθ and cotθ .

❒ Match the following :-

$\begin{tabular}{|c|c|} \cline{1-2}\bf Column A & \bf Column B \\ \cline{1-2} csc\theta & \dfrac{1}{cos\theta} \\ \cline{1-2} tan\theta & cos\theta \\ \cline{1-2} sec\theta & \dfrac{1}{sin\theta} \\ \cline{1-2} \dfrac{1}{sec\theta} & \dfrac{sin\theta}{cos\theta} \\ \cline{1-2} cot\theta & \dfrac{1}{tan\theta} \\ \cline{1-2} \end{tabular}$

$\Large{\underline{\underline{\red{\tt{\purple{\leadsto } FormulaE\:UseD:-}}}}}$

• $\orange{\boxed{\green{\tt \dag sin\theta=\dfrac{1}{csc\theta}}}}$

• $\orange{\boxed{\green{\tt \dag cos\theta=\dfrac{1}{sec\theta}}}}$

• $\orange{\boxed{\green{\tt \dag tan\theta=\dfrac{1}{cot\theta}}}}$

• $\orange{\boxed{\green{\tt \dag tan\theta=\dfrac{sin\theta}{cos\theta}}}}$

• $\orange{\boxed{\green{\tt \dag sin\theta=\dfrac{perpendicular}{hypontenuse}}}}$

• $\orange{\boxed{\green{\tt \dag cos\theta=\dfrac{base}{hypontenuse}}}}$

• $\orange{\boxed{\green{\tt \dag tan\theta=\dfrac{perpendicular}{base}}}}$

$\Large{\underline{\underline{\red{\tt{\purple{\leadsto } AnsweR:-}}}}}$

$\large\underline{\pink{\sf{\mapsto SolutioN\:\:1}}}$

Given that sinθ = 5/13 .

Now let this ratio be 5x:13x . And sinθ = $\sf{\dfrac{perpendicular}{hypontenuse}}$

So , let's find base :

$\tt:\implies hypontenuse^2=base^2+perp.^2$

$\tt:\implies 13x^2=5x^2+b^2$

$\tt:\implies 169x^2=25x^2+b^2$

$\tt:\implies b^2=169x^2-25x^2$

$\tt:\implies b^2=144x^2$

$\underline{\boxed{\red{\tt{\longmapsto \:\:base\:\:=\:\:12x}}}}$

Hence here ,

• cosθ = $\sf\dfrac{base}{hypo.}$ =12x/13x= 12/13 .
• tanθ = sinθ/cosθ = 5/12.
• cscθ = 1/sinθ = 13/5.
• secθ = 1/cosθ = 13/12.
• cotθ = 1/tanθ = 12/5 .

___________________________________

$\large\underline{\pink{\sf{\mapsto SolutioN\:\:2}}}$

(i) Given that tanθ = 15/8 .

We know cotθ = 1/tanθ = 8/15 .

(ii) Given sinθ = √3/2 .

We know cosecθ = 1/sinθ = 2/√3.

(iii) Given cosθ = 1/√2 .

We know secθ = 1/cosθ = √2/1 = √2 .

___________________________________

$\large\underline{\pink{\sf{\mapsto SolutioN\:\:3}}}$

Given that sinθ = 3/5 and cosθ = 4/5 .

We know that ,

• tanθ = sinθ/cosθ

= 3/5 ÷ 4/5

= 3/5 × 5/4

= 3/4 .

And cotθ = 1/tanθ = 1 ÷ 3/4 = 4/3.

___________________________________

$\large\underline{\pink{\sf{\mapsto MatcH\:ThE\: FollowinG:-}}}$

1) cosecθ ---------- sinθ

2) tanθ -------------- sinθ / cosθ

3) secθ -------------- 1/cosθ

4) 1/secθ ------------- cosθ

5) cotθ ---------------- 1/tanθ