Q. 5 B5.0( 2 Votes )

# Find the value of the following : (a) sin θ (b) cos θ (c) tan θ from the figures given below :  Firstly, we have to find the value of XM and we can find out with the help of Pythagoras theorem

So, In ∆XMZ

(XM)2 + (MZ)2 = (XZ)2

(XM)2 + (16)2 = (20)2

(XM)2 = (20)2 – (16)2

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

(XM)2 = (20–16)(20+16)

(XM)2 = (4)(36)

(XM)2 = 144

XM =144

XM =±12

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

Now, In ∆XMY we have the value of XM and MY but we don’t have the value of XY.

So, again we apply the Pythagoras theorem in ∆XMY

(XM)2 + (MY)2 = (XY)2

(12)2 + (5)2 = (XY)2

144 + 25 = (XY)2

(XY)2 = 169

XY =169

XY =±13

But side XY can’t be negative. So, XY = 13

a. sin θ

We know that, In ∆XMY

Side opposite to θ = MY = 5

Hypotenuse = XY = 13

So, b. cos θ

We know that, In ∆XMY

Side adjacent to θ = XM = 12

Hypotenuse = XY = 13

So c. tan θ

We know that, In ∆XMY

The side opposite to θ = MY = 5

Side adjacent to θ = XM = 12

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