Question

exercise 9.1 trigonometry all solutions

Answer 1

Solution :

Additional information :

Class - X

Board : CBSE

Exercise : 9.1

Name of the Chapter : Some Applications of trigonometry

For the Questions refer to your NCERT book :

Important things for the chapter is given as below :

(i) Student Should know the trigonometric table

(ii) Let us take a Quick view of angle of depression , line of sight and angle of elevation

Refer to Page No. 197

There are total 16 questions in the exercise 9.1 or we can say 9 chapter . I'm giving you the first four  Solution in this due to content limit . In the next questions I will do the next questions

Answer :  1

Let the pole be AB and rope be AC

Given : ∠C = 30°

To Find : Perpendicular ( AB)

We Know that ,

Sin30° = $\frac{P}{H}$

⇒ $\frac{P}{20}$

⇒ $\frac{1}{2} = \frac{p}{20}$

⇒ $\frac{20}{2}$

⇒ 10 m

Answer : AB = 10 m is the height of the pole

Answer  : 2

Given : AB ( Base ) = 8 m

AC ( Hypot) = x

BC ( Perpendicular) = y

∠C = 30°

To Find : Height of the tree

x + y = Height of the tree

Using Cos 30° ,

$\frac{B}{H}$

⇒ $\frac{\sqrt{3}}{2} * \frac{8}{AC}$

⇒ $\frac{8 * 2}{\sqrt{3} }$

⇒ $\frac{16}{\sqrt{3} }$

Using Sin 30°

P/H

⇒ 1/2 = 16/√3

⇒ 1/2 = √3/16

⇒ 16/2√3

⇒ 8/√3

Add them

16√3 + 8√3

⇒ 24√3

⇒ 8√3 ( by rationalising )

Answer : 3

let case A be below the age of 5 years and Case B be elder  

Given : Two Cases happened with a contractor

To Find : Length of the slide in each case

Case A

In ΔABC

⇒ AB (P) = 1.5m

∠ACB = 30°

Require : AC

Sin 30 = P/H

⇒ 1/2 = 1.5/AC

⇒ AC = 3 m

Case B

In ΔPQR

PQ(p) = 3 cm

∠PRQ = 60°

Require PR (H)

Sin 60°

⇒ √3/2 = 3/PR

⇒ PR = 2√3

Answer : 4

Let tower be AB and Point be C

Given : BC = 30 m

∠ACB = 30°

∠ABC = 90°

To Find : AB ( Height )

In ΔABC

Tan (C) 30° = P/B

⇒ 1/√3 = AB/30

⇒ AB = 30/√3

⇒ AB = 10√3 ( By Rationalising )

Therefore , Height of the tower is 10√3

Answer 2

Hello your answer...

Answer7

Let AB be the building and BC be the transmission tower.then,AB=20m
=>BDA=45° and CDA=60°
also let DA=Xm and
=>BC=hm
Then in right TriangleBAD,
Tan 45°=AB/DA
=>1=20/x
=>X=20
in right TriangleCAD,
Tan 60°=AC/DA=>Root over 3=20+h/x
=>X=20+h/root over3
From equation i and ii we get
=>20+h/root over 3=20
=20+h=20root over 3
=>H=20(root over3-1)m
Their for the height of the tower is 20(root over3-1)m

Answer8

=In triangleBCD
=>BC/CD=tan45°
=>BC/CD=1
=>BC=CD.....(i)
In triangleACD,
=>AB+BC/CD=root over3.......(from i)
=>1.6+BC=BCroot over3
=>BC(root over3-1)=1.6
=>BC=(1.6)(root over3+1)/(root over3-1)(root over3+1)
=>BC=1.6(root over3+1)/2=0.8(root over3+1)

Thanks,,