Evaluate the double integral. 9y2 da, d is the triangular region with vertices (0, 1), (1, 2), (4, 1) d
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Answer:
<A = 90°, <B = 71.6° <C = 18.4°
Step-by-step explanation:
Given the vertices of a △ABC to be A(−1, 6), B(2, 10), and C(7, −2)
Before we can get each angle of the triangle, we need to get its sides first.
Using the formula fie finding the distance between two points
D = √(x2-x1)²+(y2-y1)²
For side AB:
Given A(−1, 6), B(2, 10)
AB = √{2-(-1)}²+(10-6)²
AB =√3²+4²
AB = √25
AB = 5
For side AC:
Given A(-1,6) and C(7, −2)
AC = √{7-(-1)}²+(-2-6)²
AC = √8²+8²
AC = √128
AC = √64×2
AC = 8√2
For side BC:
Given B(2,10) and C(7, −2)
BC = √(7-2)²+(-2-10)²
BC = √5²+12²
BC = √25+144
BC = √169
BC = 13
First we need to get theta using cosine rule
|AC|² = |AB|+|BC| - 2|AB||AC| cos theta
(8√2)² = 5²+13²-2(5)(13)cos theta
128 = 169-130costheta
128-169 = -130costheta
-41= -130costheta
Costheta = -41/-130
Cos theta = 0.315
theta = arccos 0.315
theta = 71.64°
<B = 71.6°
Using sine rule to get <A,
BC/sin A = AC/sinB
13/sinA = 8√2/sin71.6°
8√2sinA = 13sin71.6°
SinA = 12.34/8√2
SinA = 1.09°
A = arcsin1
<A = 90.0°
<C = 180°-(71.6+90)
< C = 180-161.6
<C = 18.4°
