Each base angle is an isosceles triangle measures 55 degrees 30 minutes. Each of the congruent sides is 10 centimeters long. Estimate the following problems to the nearest tenth. A. Find the altitude of the triangle. B.what is the length of the base? C.fund the area of the triangle.
Accepted Solution
A:
Answers:
55 deg 30 Minutes = 55.5 degrees
sin = opp/hyp
sin(55.5) = altitude/10
Altitude = 10*sin(55.5)
Altitude = 10 * sin(55.5) = 8.241261886
Rounded to nearest 10th
Altitude = 8.2 cm
b. for base use Low of Cosines
Let C = base
Side A and B = length 10 Cm
c^2 = 2*a^2 -2a^2*cos(<C)
<C = 180 - 2*55.5 = 180 -111 = 69 degrees
c = sqrt ( 2a^2 - 2a^2*cos(<C)
c = sqrt ( 2*100- 200*cos(69))
c = sqrt (200 - 200*cos(69) ) = 11.32812474
Answer base c = 11.3 cms
c. Area = (1/2)*b * h Area = (1/2)*(answer a * answer b) Area = (1/2)* 11.32812474* 8.241261886 = 46.67902132
Area = 46.7 (rounded)
Area = ( 1/2) * a*b*sin(<C) = (1/2)*10*10*sin(69) = Area = 50* sin(69) = 46.67902132 Area = 46.7 (rounded)