Question

16. Find all missing sides and angles for triangle ABC, where a,b, and c are lengths of the sides opposite angles A, B, andC respectively. Round the answers to the nearest tenth.a=3 ft., b = 6 ft., c= 7 ft.A= degrees B= degrees C= degrees

Answer 1

We have

We will use the Law of cosines in order to find the angle C

`c^2=a^2+b^2-2ab\cdot cos\mleft(C\mright)`

we need to isolate C

`\cos (C)=(c^2-a^2-b^2)/(-2ab)`

where

a=3 ft

b = 6 ft

c= 7 ft.

we substitute the values

`\cos (C)=(7^2-3^2-6^2)/(-2(3)(6))=-(1)/(9)`
`C=\cos ^(-1)(-(1)/(9))=96.38\text{\degree}`

Then we will use the law of sines

`(\sin(96.38))/(7)=(\sin(A))/(3)`
`\sin (A)=(3\cdot\sin (96.38))/(7)`
`\sin (A)=0.426`
`A=\sin ^(-1)(0.426)=25.21`

Then we use the fact that the sum of the interior angles of a triangle is 180°

A+B+C=180

B=180-25.21-96.38=58.41°

The solution is

A= 25.21 degrees

B= 58.41 degrees

C= 96.38 degrees