Each base angle of an isosceles triangle measures 55° 30'. Each of the congruent sides is 10 centimeters long. Estimate the following problems to the nearest tenth. a. Find what altitude of the triangle
b. What is the length of the base?
c. Find the area of the triangle

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)

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