the right angle triangle ABC,angleB=90° ,if tan C= √3 the value of the angle A is
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In a Δ ABC, right angled at A, if tanC =
3
, find the value of sinBcosC+cosBsinC.
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Answer
In right angled Δ ABC,
Given, tanC=
3
We know that, tanθ=
adjacent side to∠θ
opposite side to∠θ
∴AB=
3
and AC=1
From Pythagoras theorem,
BC
2
=AB
2
+AC
2
BC
2
=(
3
)
2
+1
2
BC
2
=3+1=4
BC=2
sinθ=
hypotenuse
opposite side to∠θ
cosθ=
hypotenuse
adjacent side to∠θ
sinB=
BC
AC
=
2
1
cosB=
BC
AB
=
2
3
sinC=
BC
AB
=
2
3
cosC=
BC
AC
=
2
1
Therefore,
sinBcosC + cosBsinC
=(
2
1
)(
2
1
) + (
2
3
)(
2
3
)
=
4
1
+
4
3
=
4
4
=1
Step-by-step explanation:
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Answer:
1
Step-by-step explanation:
In right angled Δ ABC,
Given, tanC=
3
We know that, tanθ=
adjacent side to∠θ
opposite side to∠θ
∴AB=
3
and AC=1
From Pythagoras theorem,
BC
2
=AB
2
+AC
2
BC
2
=(
3
)
2
+1
2
BC
2
=3+1=4
BC=2
sinθ=
hypotenuse
opposite side to∠θ
cosθ=
hypotenuse
adjacent side to∠θ
sinB=
BC
AC
=
2
1
cosB=
BC
AB
=
2
3
sinC=
BC
AB
=
2
3
cosC=
BC
AC
=
2
1
Therefore,
sinBcosC + cosBsinC
=(
2
1
)(
2
1
) + (
2
3
)(
2
3
)
=
4
1
+
4
3
=
4
4
=1