Problem 2 Consider a triangle ABC with the angles 0 Solution

Problem #2: Consider a triangle ABC with the angles 0

Solution

Problem#2:

given =cos^-1(3_/5) =>cos = (3_/5)

=>sin =(((5²-3²))_/5)

=>sin =(4_/5)

=sin^-1(2_/5)

=>sin=(2_/5)

size of side opposite to =a= 5 cm ,size of side opposite to =b,size of side opposite to =c

by law of sines a_/sin=b_/sin=c_/sin

=>5_/(4/5)=b_/sin=c_/(2/5)

=>c=(5_/(4/5))*(2_/5)

=>c=((55)_/2)

=sin^-1(4_/5),=sin^-1(2_/5) ,++=180^o

=> =180^o-sin^-1(4_/5) -sin^-1(2_/5)

=> =63.435^o

area of triangle ABC= (1_/2)ac sin

area of triangle ABC= (1_/2)*5*((55)_/2)* sin(63.435^o)

area of triangle ABC= 12.5 cm²

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