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3 years ago

Find the measures of the exterior angles.

Mathematics

1 answer:

Nesterboy [21]3 years ago

8 0

Answer:

The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles. (Non-adjacent interior angles may also be referred to as remote interior angles.) FACTS: An exterior ∠ is equal to the addition of the two Δ angles not right next to it.

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Match each value with its formula for ABC.

MariettaO [177]

The solution to the question is:

c is 6 = $\sqrt{a^{2} + b^{2} -2abcosC }$

b is 5 = $\sqrt{a^{2} + c^{2} -2accosB }$

cosB is 2 = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

a is 4 = $\sqrt{b^{2} + c^{2} -2bccosA }$

cosA is 3 = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

cosC is 1 = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

<h3>What is cosine rule?</h3>

it is used to relate the three sides of a triangle with the angle facing one of its sides.

The square of the side facing the included angle is equal to the some of the squares of the other sides and the product of twice the other two sides and the cosine of the included angle.

Analysis:

If c is the side facing the included angle C, then

$c^{2}$ = $a^{2}$ + $b^{2}$ -2ab cos C-----------------1

then c = $\sqrt{a^{2} + b^{2} -2abcosC }$

if b is the side facing the included angle B, then

$b^{2}$ = $a^{2}$ + $c^{2}$ -2accosB-----------------2

b = $\sqrt{a^{2} + c^{2} -2accosB }$

from equation 2, make cosB the subject of equation

2ac cosB = $a^{2}$ + $c^{2}$ - $b^{2}$

cosB = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

if a is the side facing the included angle A, then

$a^{2}$ = $b^{2}$ + $c^{2}$ -2bccosA--------------------3

a = $\sqrt{b^{2} + c^{2} -2bccosA }$

from equation 3, making cosA subject of the equation

2bcosA = $b^{2}$ + $c^{2}$ - $a^{2}$

cosA = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

from equation 1, making cos C the subject

2abcosC = $b^{2}$ + $a^{2}$ - $c^{2}$

cos C = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

In conclusion,

c is 6 = $\sqrt{a^{2} + b^{2} -2abcosC }$

b is 5 = $\sqrt{a^{2} + c^{2} -2accosB }$

cosB is 2 = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

a is 4 = $\sqrt{b^{2} + c^{2} -2bccosA }$

cosA is 3 = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

cosC is 1 = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

Learn more about cosine rule: brainly.com/question/4372174

$SPJ1

4 0

2 years ago

Someone help me.......

Makovka662 [10]

The tan of a is 12/5

3 0

3 years ago

In a rhombus whose side measures 20 and the smaller angle is 38°. Find the length of the larger diagonal, to the nearest tenth.

Lemur [1.5K]

A rhombus diagonals form right angles. The diagonals also cut in half the angles

use cosine: cos38 = $\frac{a}{20\\\\}$

a = 20cos38

a=15.8

4 0

3 years ago

Please help I appreciate it

Anni [7]

Answer:

Step-by-step explanation:

g(x) got smaller

A - makes it thinner

B & C - move it to the left or right

D- makes it wider

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3 years ago

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A cylinder has a height of 16 cm and a radius of 5 cm. A cone has a height of 12 cm and a radius of 4 cm. If the cone is placed

AlexFokin [52]

Given:
Cylinder: height = 16 cm ; radius = 5 cm
cone: height = 12 cm ; radius = 4 cm

Volume of cylinder = 3.14 * (5cm)² * 16cm = 1,256 cm³
Volume of cone = 3.14 * (4cm)² * 12cm/3 = 200.96 cm³

Volume of air space = 1256 cm³ - 200.96 cm³ = 1,055.04 cm³

4 0

3 years ago

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