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3 years ago

PLZ HELP !!!!!! ASAP!!!

Mathematics

2 answers:

Ad libitum [116K]3 years ago

8 0

Answer & Step-by-step explanation:

(a)

The hypotenuse is on line CA (the hypotenuse is always opposite the 90° angle (marked by a little square))

The adjacent is on the line BA (adjacent is next to the given angle, but NOT the hypotenuse)

The opposite is on the line CB (this is opposite the given angle)

(b)

i. false (b is a right angle)

ii. false (the side opposite C is BA)

iii. true

iv. false (the side opposite B is the hypotenuse, and the hypotenuse is always the longest side in a triangle)

(c)

cosine ratio: $cos=\frac{adjacent}{hypotenuse}$

tangent ratio: $tan=\frac{opposite}{adjacent}$

The cosine and tangent ratios of the given angle:

$cos0=\frac{AB}{CA} \\\\tan0=\frac{CB}{AB}$

(d)

Remember SOH-CAH-TOA:

Sine=Opposite/Hypotenuse

Cosine=Adjacent/Hypotenuse

Tangent=Opposite/Adjacent

Using the angle C, plug in the appropriate sides:

$sinC=\frac{BA}{CA}\\\\ cosC=\frac{CB}{CA}\\\\ tanC=\frac{BA}{CB}$

:Done

Brrunno [24]3 years ago

7 0

Part (a)

BC = opposite side (furthest leg from the reference angle)

AB = adjacent side (closest leg from the reference angle)

AC = hypotenuse (always opposite the 90 degree angle)

=============================================

Part (b)

i. False. Angle B is 90 degrees as shown by the square angle marker.

ii. False. Side AB is opposite angle C. Note how "C" is part of "BC", so that means we cannot have BC be opposite C.

iii. True. Leg AB is the closer leg to angle A. We have "A" in "AB" to see this without having to draw the diagram. Refer to part (a) above.

iv. False. The longest side of any right triangle is always the hypotenuse. The longest side of any triangle is always opposite the largest angle.

==============================================

Part (c)

cos(theta) = adjacent/hypotenuse = AB/AC

tan(theta) = opposite/adjacent = BC/AB

Refer back to part (a) to determine the opposite,adjacent and hypotenuse side lengths.

==============================================

Part (d)

The reference angle has changed, so the opposite and adjacent sides swap. The hypotenuse remains the same regardless of what reference angle you pick.

sin(C) = opposite/hypotenuse = AB/AC

cos(C) = adjacent/hypotenuse = BC/AC

tan(C) = opposite/adjacent = AB/BC

Note the tangent ratio is the reciprocal of what we found back in part (c).

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Answer:

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Eight less than

The difference of a number and five

Is twelve.

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Verify sine law by taking triangle in 4 quadrant<br>Explain with figure.<br>

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Proof of the Law of Sines

The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)

 sin  A

 sin  B

 sin  C

For more see Law of Sines.

Acute triangles

Draw the altitude h from the vertex A of the triangle

From the definition of the sine function

 sin  B =

a n d  sin  C =

h = c  sin  B a n d h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

 sin  C

 sin  B

Repeat the above, this time with the altitude drawn from point B

Using a similar method it can be shown that in this case

 sin  C

 sin  A

Combining (4) and (5) :

 sin  A

 sin  B

 sin  C

- Q.E.D

Obtuse Triangles

The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.

First the interior altitude. This is the same as the proof for acute triangles above.

Draw the altitude h from the vertex A of the triangle

 sin  B =

a n d  sin  C =

h = c  sin  B a n d h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

 sin  C

 sin  B

Draw the second altitude h from B. This requires extending the side b:

The angles BAC and BAK are supplementary, so the sine of both are the same.

(see Supplementary angles trig identities)

Angle A is BAC, so

 sin  A =

h = c  sin  A

In the larger triangle CBK

 sin  C =

h = a  sin  C

From (6) and (7) since they are both equal to h

c  sin  A = a  sin  C

Dividing through by sinA then sinC:

 sin  A

 sin  C

Combining (4) and (9):

 sin  A

 sin  B

 sin  C

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