In triangle ABC, AB=8cm, angle A =50degree, angle B=70 degree. What is the perpendicular distance from C to AB.also find area of the triangle?
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Step-by-step explanation:
Let CD be perpendicular from C to AB.
(a) In ΔACD,
sin50
o
=
AC
CD
⇒0.7660=
5
CD
⇒CD=0.7660×5=3.83cm
(b) cos50
o
=
AC
AD
⇒0.6428=
5
AD
⇒AD=0.6428×5=3.214
DB = AB - AD = (8 - 3.214) cm = 4.786 cm
In right triangle CDB,
BC
2
=CD
2
+DB
2
=(3.83)
2
+(4.786)
2
=14.67+22.91
=37.58
⇒BC=
37.58cm
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