Question

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

Answer 1

Answer:

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.

Step-by-step 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 $c = AB = 6$.

$\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}$.

$\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}$.

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

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:

$a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}$.

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

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.

$b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}$.

$\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}$.

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

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:

$\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}$

$\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9$.