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

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,