@Genuisusers
@Moderators
@bestusers
In a right triangle ABC , right angled at B , if tan A = 1 . Find 2 sin a * cos a . (trigonometry)
• No Spam
Answers
Step-by-step explanation:
In △ABC, ∠ABC=90
o
∴tanA=
AB
BC
Since tanA=1 (Given)
AB
BC
=1 ∴BC=AB
Let AB=BC=k, where k is a positive number.
Now, AC
2
=AB
2
+BC
2
∴AC=
AB
2
+BC
2
=
k
2
+k
2
∴AC=k
2
∴sinA=
AC
BC
=
k
2
k
=
2
1
,cosA=
AC
AB
=
k
2
k
=
2
1
2sinAcosA=2(
2
1
)(
2
1
)=1
∴2sinAcosA=1
Answer:
✨✨Your answer✨✨
In right triangle ABC
Angle C=90 degree
tan A=1
To verify that
2sin Acos A=1
Solution:
tan A=1
tan A=\frac{Perpendicular\;side}{Base}=\frac{1}{1}
BC=k
AC=k
Using Pythagoras theorem
(Hypotenuse)^2=(Perpendicular)^2+(Base)^2
AB^2=AC^2+BC^2
AB^2=k^2+k^2=2k^2
AB=\sqrt{2k^2}=k\sqrt{2}
Sin A=\frac{Perpendicular\;side}{Hypotenuse}=\frac{BC}{AB}=\frac{k}{k\sqrt{2}}=\frac{1}{\sqrt{2}}
cos A=\frac{Base}{Hypotenuse}=\frac{AC}{AB}=\frac{k}{k\sqrt{2}}=\frac{1}{\sqrt{2}}
Substitute the values
2sinAcos A=2\times \frac{1}{\sqr{2}}\times \frac{1}{\sqrt{2}}
2sinA cos A=\frac{2}{(\sqrt{2})^2}=\frac{2}{2}
2sinA cos A=1
Hence, verified.
Step-by-step explanation:
Hope it will help✨✨