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3 years ago

If measure angle XAB=32 What is measure angle CAX?

Mathematics

2 answers:

blagie [28]3 years ago

8 0

I believe the answer would be 64.

Svetradugi [14.3K]3 years ago

6 0

The measure of CAX is 148. You subtract how many degrees the straight is(180 degrees) by the angle of which you know how many degrees it is(32 degrees). 180-32= 148 degrees

I hope this helps!

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Help please!<br> Find perimeter and Area

MatroZZZ [7]

Answer:

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Step-by-step explanation:

Sides:

a = 29 m

b = 20 m

c = 21 m

Angles:

A = 90 °

B = 43.6028 °

C = 46.3972 °

Other:

P = 70 m

s = 35 m

K = 210 mi^2

r = 6 m

R = 14.5 m

Agenda:

A = angle A

B = angle B

C = angle C

a = side a

b = side b

c = side c

P = perimeter

s = semi-perimeter

K = area

r = radius of inscribed circle

R = radius of circumscribed circle

SSS is Side, Side, Side

Heron’s formula says that if a triangle ABC has sides of lengths a, b, and c opposite the respective angles, and you let the semiperimeter, s, represent half of the triangle’s perimeter, then the area of the triangle is

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2 years ago

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MariettaO [177]

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Step-by-step explanation:

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3 years ago

A sprinkler system uses 800 gallons of water in 5 days. How many gallons will the system use in a week?

emmainna [20.7K]

Yes 1,120 gallons per week (7 days)
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3 years ago

Find the EXACT value of sin(A−B) if cos A = 3/5 where A is in Quadrant IV and cos B = 12/13 where B is in Quadrant IV. Assume al

MissTica

$\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)$

well, for both angles A and B we're on the IV Quadrant, meaning, the sine is negative, the cosine is positive, likewise, the opposite side is negative and the adjacent side for the angle is positive.

$\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5$

$\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}$

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3 years ago