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3 years ago

What is the measure of A? And B

Mathematics

2 answers:

sladkih [1.3K]3 years ago

7 0

Answer:

A = 72°

B = 108°

Step-by-step explanation:

5y - 3 = 3y + 27

5y - 3y = 27 + 3

2y = 30

y = 30/2

y = 15

A = 5y - 3

A = 5(15) - 3

A = 75 - 3

A = 72°

A + B = 180°

72° + B = 180°

B = 180° - 72°

B = 108°

bekas [8.4K]3 years ago

6 0

A=72
B=108
That’s should be the answer

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zmey [24]

Answer:

Part 1) $x=\frac{15}{2}\ units$

Part 2) $z=\frac{15\sqrt{3}}{2}\ units$

Part 3) $y= \frac{15\sqrt{3}}{4}\ units$

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Step-by-step explanation:

step 1

Find the value of x

In the large right triangle

$cos(60^o)=\frac{x}{15}$ ----> by CAH (adjacent side divided by the hypotenuse)

Remember that

$cos(60^o)=\frac{1}{2}$

substitute

$\frac{1}{2}=\frac{x}{15}$

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step 2

Find the value of z

In the large right triangle

Applying the Pythagorean Theorem

$15^2=x^2+z^2$

substitute the value of x

$15^2=(\frac{15}{2})^2+z^2$

solve for z

$z^2=15^2-(\frac{15}{2})^2$

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step 3

Find the value of y

In the right triangle of the right

$sin(30^o)=\frac{y}{z}$ ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values of y and z

Remember that

$sin(30^o)=\frac{1}{2}$

$\frac{1}{2}=y:\frac{15\sqrt{3}}{2}$

solve for y

$\frac{1}{2}= \frac{2y}{15\sqrt{3}}$

$y= \frac{15\sqrt{3}}{4}\ units$

step 4

Find the value of b

In the right triangle of the right

$cos(30^o)=\frac{b}{z}$ ---> by CAH (adjacent side divided by the hypotenuse)

substitute the given values of y and z

Remember that

$cos(30^o)=\frac{\sqrt{3}}{2}$

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solve for y

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Mumz [18]

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