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.

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)