Q. 5 A3.7( 3 Votes )

# Find the value of the following : (a) sin θ (b) cos θ (c) tan θ from the figures given below :

Answer :

Firstly, we give the name to the midpoint of BC i.e. M

BC = BM + MC = 2BM or 2MC

BM = 5 and MC = 5

Now, we have to find the value of AM, and we can find out with the help of Pythagoras theorem.

So, In ∆AMB

(AM)2 + (BM)2 = (AB)2

(AM)2 + (5)2 = (13)2

(AM)2 = (13)2 – (5)2

Using the identity a2 –b2 = (a+b) (a – b)

(AM)2 = (13–5)(13+5)

(AM)2 = (8)(18)

(AM)2 = 144

AM =144

AM =±12

But side AM can’t be negative. So, AM = 12

a. sin θ

We know that,

In ∆AMB

Side opposite to θ = AM = 12

Hypotenuse = AB=13

So,

So,

b. cos θ

We know that,

In ∆AMB

The side adjacent to θ = BM = 5

Hypotenuse = AB = 13

So,

So,

c. tan θ

We know that,

In ∆AMB

Side opposite to θ = AM = 12

The side adjacent to θ = BM = 5

So,

So,

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