if
AB=3
EB=7
CA=4
then AC=?
Answers
Answered by
1
Answer:
Explanation:
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
.
Answered by
0
By Pythagoras Theorem,
⇒ AC2 = AB2+ BC2
Given in ΔABC, AB = 3, BC = 4, AC = 5.
⇒ 52 = 32 + 42
⇒ 25 = 25
∴ ΔABC is a right angled triangle and ∠B is a right angle.
We know that the radius of the circle touching all the sides is (AB + BC – AC )/ 2
⇒ The required radius of circle = (3 + 4 – 5 )/2 = 2/2 = 1
I hope you get your answer
⇒ AC2 = AB2+ BC2
Given in ΔABC, AB = 3, BC = 4, AC = 5.
⇒ 52 = 32 + 42
⇒ 25 = 25
∴ ΔABC is a right angled triangle and ∠B is a right angle.
We know that the radius of the circle touching all the sides is (AB + BC – AC )/ 2
⇒ The required radius of circle = (3 + 4 – 5 )/2 = 2/2 = 1
I hope you get your answer
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