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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