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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Answer:
Length = 15
Width = 6
Area = 90
Step-by-step explanation:
Area of the rectangle =
Lets factorise the polynomial
We know that Area = Length × Width
Let the length be (x + 3) & width be (x - 6)
Its given in the question that x = 12. So putting the value of x gives ,
Length =
Width =
Area = Length × Width = 15 × 6 = 90