TrangleABC IN KONEB IS RIGHT ANGLE TRNGLE AND BC =5 THETA =30° FIND AB AND AC
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triangle ABC right angled at B theta=30 =x
tanx=AB/BC
tan30=AB/5
5/root 3=AB
by PYT therom
AC^2=25/3+5
AC=root40/root3
tanx=AB/BC
tan30=AB/5
5/root 3=AB
by PYT therom
AC^2=25/3+5
AC=root40/root3
Answered by
0
Hi ,
In Triangle ABC , <B = 90°
BC = 5
< BAC = theta = 30°
i ) cot 30 ° = AB / BC
√3 = AB / 5
5√3 = AB
AB = 5√3
ii ) sin 30° = BC / AC
1/2 = 5 / AC
AC = 5 × 2
AC = 10
Therefore ,
AB = 5√3
AC = 10
I hope this helps you.
:)
In Triangle ABC , <B = 90°
BC = 5
< BAC = theta = 30°
i ) cot 30 ° = AB / BC
√3 = AB / 5
5√3 = AB
AB = 5√3
ii ) sin 30° = BC / AC
1/2 = 5 / AC
AC = 5 × 2
AC = 10
Therefore ,
AB = 5√3
AC = 10
I hope this helps you.
:)
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