in triangle ABC angle B is 90 , AB=5 & BC +AC=25 find the value of sin a, cos a, tan a
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In the given figure, ΔABC circumscribed the circle with centre O.
Radius OD=3cm
BD=6cm,DC=9cm
Area of ΔABC=54cm2
To find : Length of AB and AC.
AF and EA are tangents to the circle at point A.
Let AF=EA=x
BD and BF are tangents to the circle at point B.
BD=BF=6cm
CD and CE are tangents to the circle at point C.
CD=CE=9cm
Now, new sides of the triangle are:
AB=AF+FB=x+6cm
AC=AE+EC=x+9cm
BC=BD+DC=6+9=15cm
Now, using Heron's formula:
Area of triangle ABC=s(s−a)(s−b)(s−c)
Where S=2a+b+c
S=1/2(x+6+x+9+15)=x+15
Area of ABC=(x+15)(x+15−(x+6))(x+15−(x−9))(x+15−15)
Or
54=(x+15)(9)(6)(x)
Squaring both sides, we have
Step-by-step explanation:
hopes its helps
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