Each base angle is an isosceles triangle measures 55 degrees 30 minutes. Each of the congruent sides is 10 centimeters long. Est imate 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.
1 answer:
Answers: <span> 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)</span>
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