Step-by-step explanation:
problem b
isosceles triangle 2 angles are equal.
180-38 = 142
each one = 71
draw a vertical line bisecting the base 8cm. it will be the ht of triangle.
area = .5*b*h
the vertical line will make 90 with base.
tan 71 = h/4
h = 4 * 2.9=11.61
area = 11.61*8*.5=46.46
problem c is interesting
let's find hypotenuse =
[tex] \sqrt{ {6.1 }^{2} + {4.3}^{2} } [/tex]
= 7.46
sin a = 4.3/7.46
a = 34.75
angle a is alternate with the angle made by 4.3 on the upper parallel line. line 4.3 is a transversal to the parallel lines. also draw an imaginary line from the pt, 4.3 meets the lower parallel line to the upper parallel line. now yu will get a right triangle with 4.3 as hypotenuse.
the imaginary line is the dist between 2 parallel lines.
let it be b
b/ 4.3 = sin 34.75
b=4.3*sin34.75 = 2.47
2.47 cm is the dist btw parallel lines
