In triangle ABC, AB=c, BC=a, CA=b
1 answer:
Answer:
Step-by-step explanation:
Correct option is
B
AE
2
=
9
73
C
tanθ=
8
3
Hence
BD=DE=EC=
3
5
and △ABC is a right angled triangle.
Now
cosC=
2.AC.EC
AC
2
+EC
2
−AE
2
Or
AE
2
=AC
2
+EC
2
−2.AC.EC.cosC
Or
=4
2
+(
3
5
)
2
−2.8.
3
5
.cosC
=16+
9
25
−
3
2×4×5
.
5
4
Or
=
9
169
−
3
32
=
9
169−96
=
9
73
.
Hence
AE
2
=
9
73
.
cosθ=
2.AE.4
AE
2
+4
2
−
3
5
2
=
73
8
.
Hence
secθ=
8
73
Now
tanθ=
sec
2
θ−1
=
64
73
−1
=
8
3
.
solution
Expand-image
Correct option is
B
AE
2
=
9
73
C
tanθ=
8
3
Hence
BD=DE=EC=
3
5
and △ABC is a right angled triangle.
Now
cosC=
2.AC.EC
AC
2
+EC
2
−AE
2
Or
AE
2
=AC
2
+EC
2
−2.AC.EC.cosC
Or
=4
2
+(
3
5
)
2
−2.8.
3
5
.cosC
=16+
9
25
−
3
2×4×5
.
5
4
Or
=
9
169
−
3
32
=
9
169−96
=
9
73
.
Hence
AE
2
=
9
73
.
cosθ=
2.AE.4
AE
2
+4
2
−
3
5
2
=
73
8
.
Hence
secθ=
8
73
Now
tanθ=
sec
2
θ−1
=
64
73
−1
=
8
3
solution
Expand-image
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