Answer:
A rectangle has dimensions 10 cm by 6 cm. determine the measures of the angles at the point of where the diagonals intersect
1 answer:
See the attached figure to better understand the problem
we know that
in the triangle ABC
tan(alfa/2)=AC/BC----> 5/3
so
(alfa/2)=arctan (5/3)---------> 59.0362°
alfa=2*59.0362--------> alfa=118.07°
alfa+beta=180°----------> because are supplementary angles
beta=180-118.07-------> beta=61.93°
the answer is
the angles are
alfa=118.07°
beta=61.93°
we know that
in the triangle ABC
tan(alfa/2)=AC/BC----> 5/3
so
(alfa/2)=arctan (5/3)---------> 59.0362°
alfa=2*59.0362--------> alfa=118.07°
alfa+beta=180°----------> because are supplementary angles
beta=180-118.07-------> beta=61.93°
the answer is
the angles are
alfa=118.07°
beta=61.93°
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Step-by-step explanation:
A ball is launched straight up in the air from a height of 6 feet. The velocity as a function of time t is given by :
f(t) = -32 t+285
Height of the ball is :
C is constant. Here the ball is launched from a height of 6 feet. So,
At t = 2 s,
At t = 9 s,
Between 2 s and 9 s, the ball's height changed is : 1275 - 512 = 763 m.