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Question
In the adjoining figure, `∠B = 90° , ∠BAC = theta° , BC = CD = 4cm and AD = 10 cm`. find (i) sin theta and (ii) `costheta`
Solution
In ΔABD,
Using Pythagoras theorem, we get
AB= `sqrt(AD^2-BD^2)`
= `sqrt(10^2-8^2)`
=`sqrt(100-64)`
=`sqrt(36)`
=6cm
Again,
In ΔABC,
Using Pythagoras therem, we get
AC= `sqrt(AB^2 +BC^2)`
=`sqrt(6^2+4^2)`
=`sqrt(36+16)`
=`sqrt(52)`
=2`sqrt(13)`cm
Now,
(i) `sintheta = (BC)/(AC)`
=`4/(2sqrt(13))`
=`2/sqrt(13)`
=`(2 sqrt(13))/13`
(ii) `cos theta = (AB)/(AC)`
= `6/(2sqrt(13))`
=`3/sqrt(13)`
=`(3sqrt(13))/13`
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