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3 years ago

I need help finding the measures of the other angles.

Mathematics

2 answers:

Alex Ar [27]3 years ago

5 0

<7 and <10 are each 84 degrees

solniwko [45]3 years ago

5 0

We know that Sum of Angles in a Triangle = 180

⇒ Angle(4) + Angle(5) + Angle(7) = 180

⇒ 52 + 44 + Angle(7) = 180

⇒ 96 + Angle(7) = 180

⇒ Angle(7) = 180 - 96 = 84

We know that Vertically Opposite Angles are Equal

Angle(7) and Angle(10) are Equal

⇒ Angle(10) = 84

Also, Angle(8) and Angle(9) are Equal

Let us take Angle(8) = Angle(9) = P

As Angle(8) - Angle(9) - Angle(7) - Angle(10) form a Complete Angle, their Sum should be Equal to 360.

⇒ Angle(8) + Angle(9) + Angle(7) + Angle(10) = 360

⇒ P + P + 84 + 84 = 360

⇒ 2P + 168 = 360

⇒ 2P = 192

⇒ P = 96

We took Angle(8) = Angle(9) = P

⇒ Angle(8) = Angle(9) = 96

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

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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:

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.

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:

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

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

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.

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

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

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