Question

Please dont ignore, Need help!!! Use the law of sines/cosines to find..

Answer 1

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Explanation:

16

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

• 
  `sin(A) = \sin{103\textdegree{}}`,
• The opposite side of angle A
  `a = BC = 26`,
• The angle C is to be found, and
• The length of the side opposite to angle C
  `c = AB = 6`.

`\displaystyle (sin(C))/(sin(A)) = (c)/(a)`.

`\displaystyle sin(C) = (c)/(a)\cdot sin(A) = (6)/(26)* sin(103\textdegree)`.

`\displaystyle C = \sin^(-1){(sin(C)}) = \sin^(-1){\left((c)/(a)\cdot sin(A)\right)} = \sin^(-1){\left((6)/(26)* sin(103\textdegree)}\right)} = 13.0\textdegree{}`.

Note that the inverse sine function here
`\sin^(-1)()` is also known as arcsin.

17

By the law of cosine,

`c^(2) = a^(2) + b^(2) - 2\;a\cdot b\cdot cos(C)`,

where

• 
  `a`,
  `b`, and
  `c` are the lengths of sides of triangle ABC, and
• 
  `cos(C)` is the cosine of angle C.

For triangle ABC:

• 
  `b = 21`,
• 
  `c = 30`,
• The length of
  `a` (segment BC) is to be found, and
• The cosine of angle A is
  `cos(123\textdegree)`.

Therefore, replace C in the equation with A, and the law of cosine will become:

`a^(2) = b^(2) + c^(2) - 2\;b\cdot c\cdot cos(A)`.

`\displaystyle \begin{aligned}a &= \sqrt{b^(2) + c^(2) - 2\;b\cdot c\cdot cos(A)}\\&=\sqrt{21^(2) + 30^(2) - 2* 21* 30 * cos(123\textdegree)}\\&=45.0 \end{aligned}`.

18

For triangle ABC:

• 
  `a = 14`,
• 
  `b = 9`,
• 
  `c = 6`, and
• Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

`b^(2) = a^(2) + c^(2) - 2\;a\cdot c\cdot cos(B)`.

`\displaystyle cos(B) = (a^(2) + c^(2) - b^(2))/(2\;a\cdot c)`.

`\displaystyle B = \cos^(-1){\left((a^(2) + c^(2) - b^(2))/(2\;a\cdot c)\right)} = \cos^(-1){\left((14^(2) + 6^(2) - 9^(2))/(2* 14* 6)\right)} = 26.0\textdegree`.

15

For triangle DEF:

• The length of segment DF is to be found,
• The length of segment EF is 9,
• The sine of angle E is
  `sin(64\textdegree)}`, and
• The sine of angle D is
  `sin(39\textdegree)`.

Apply the law of sine:

`\displaystyle (DF)/(EF) = (sin(E))/(sin(D))`

`\displaystyle DF = (sin(E))/(sin(D))\cdot EF = (sin(64\textdegree))/(39\textdegree) * 9 = 12.9`.