NOT TO SCALE all values in feet For the triangle above side

(NOT TO SCALE, all values in feet)

For the triangle above, side AC is 111 ft long, side BC is 160 ft long, and angle ACB is 106°.

A.)What is the length of AB? (ft)

D.)What is the area of the triangle ABC? (sq ft)

Solution

We have given side AC=111ft,side BC=160ft and angle ACB is 106°

Let AC=b, BC=a and AB=c are sides of the triangle

By using cosine rule

c²=a²+b²-2ab*cosC

c²=(160)²+(111)²-2(160)*(111)*cos(106)

c²=(160)²+(111)²-35520*(-0.27)=47511.4

c=sqrt(47511.4)=217.9ft

A) The length of AB=c=217.9 ft

Triangle ADC is right triangle and we have angles <D=90 and <C/2=53

From Triangle ADC <A+<D+<C/2 =180 implies <A=180-90-53=37

Triangle CDB is right triangle and we have angles <D=90 and <C/2=53

From Triangle CDB <C/2+<D+<B =180 implies <B=180-90-53=37⁰

B)The angle of ABC=37⁰

We know the triangle area formula is sqrt(s(s-a)(s-b)(s-c))

a,b,c are sides and s=(a+b+c)/2 =(160+111+217.9)/2=244.45

triangle area is

sqrt(244.45*(244.45-160)*(244.45-111)*(244.45-217.9))=8552.36

Therefore

D) The area of triangle of ABC=8552.36 sqft

C) To get CD we have The area of triangle of ABC =1/2 base *height =(1/2)*AB*CD

The area of triangle of ABC =1/2 base *height =(1/2)*AB*CD=(1/2)*(217.9)*CD

The area of triangle of ABC=(1/2)*(217.9)*CD=8552.36  

since from above The area of triangle of ABC=8552.36 sqft

(1/2)*(217.9)*CD=8552.36

CD=(8552.36 *2)/(217.9)=78.49ft

The length of CD is 78.49ft