Mathematics, 21.12.2019 19:31 carcon2019

1. consider the right triangle abc given below.
a. find the length of side b to two decimal places.
b. find the length of side a to two decimal places in three different ways.

2. solve the triangles below.

1. consider the right triangle abc given below. a. find the length of side b to two decimal p

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Answer from: cathydaves

Part 1)

Part a) $b=10.57\ units$

Part b) $a=22.66\ units$ (three different ways in the procedure)

Part 2)

First triangle (triangle a)

Part a) $c=1.35\ units$

Part b) $A=35.11\°$

Part c) $B=129.89\°$

Second triangle (triangle b)

Part a) $A=83\°$

Part b) $AC=10.77\ units$

Part c) $BC=15.11\ units$

Step-by-step explanation:

Part 1)

Part A

we know that

In the right triangle ABC

$sin(B)=\frac{AC}{AB}$

we have

$B=25\°$

$AC=b\ units$

$AB=25\ units$

Substitute and solve for b

$sin(25\°)=\frac{b}{25}$

$b=25*sin(25\°)=10.57\ units$

Part B

First way

we know that

In the right triangle ABC

$cos(B)=\frac{BC}{AB}$

we have

$B=25\°$

$BC=a\ units$

$AB=25\ units$

Substitute and solve for a

$cos(25\°)=\frac{a}{25}$

$a=25*cos(25\°)=22.66\ units$

Second way

Applying the Pythagoras theorem

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

we have

$c=25\ units$

$b=10.57\ units$

substitute and solve for a

$25^{2}=a^{2} +10.57^{2}$

$a^{2}=25^{2}-10.57^{2}$

$a=22.66\ units[$

Third way

we know that

In the right triangle ABC

$tan(B)=\frac{AC}{BC}$

we have

$B=25\°$

$BC=a\ units$

$AC=b=10.57\ units$

substitute and solve for a

$tan(25\°)=\frac{10.57}{a}\\ \\a=10.57/ tan(25\°)\\ \\a=22.66\ units$

Part 2)

triangle a

we have

$C=15\°$

$a=3\ units$

$b=4\ units$

Step 1

Find the measure of length side c

Applying the law of cosines

$c^{2}=a^{2}+b^{2}-2(a)(b)cos(C)$

substitute

$c^{2}=3^{2}+4^{2}-2(3)(4)cos(15\°)$

$c^{2}=25-24cos(15\°)$

$c=1.35\ units$

Step 2

Find the measure of angle A

Applying the law of sines

$\frac{a}{sin(A)}=\frac{c}{sin(C)}$

we have

$a=3\ units$

$c=1.35\ units$

$C=15\°$

substitute and solve for A

$\frac{3}{sin(A)}=\frac{1.35}{sin(15\°)}\\ \\sin(A)=3*sin(15\°)/1.35\\ \\sin(A)=0.5752\\ \\A=35.11\°$

Step 3

Find the measure of angle B

Remember that the sum of the internal angles of a triangle must be equal to 180 degrees

$A+B+C=180\°$

we have

$C=15\°$

$A=35.11\°$

substitute

$B=180\°-35.11\°-15\°=129.89\°$

triangle b

we have

$C=52\°$

$B=45\°$

$c=12\ units$

Step 1

Find the measure of angle A

Remember that the sum of the internal angles of a triangle must be equal to 180 degrees

$A+B+C=180\°$

we have

$C=52\°$

$B=45\°$

substitute

$A=180\°-52\°-45\°=83\°$

Step 2

Find the measure of side AC

Applying the law of sines

$\frac{b}{sin(B)}=\frac{c}{sin(C)}$

we have

b=AC

$c=12\ units$

$B=45\°$

$C=52\°$

substitute and solve for b

$\frac{b}{sin(45\°)}=\frac{12}{sin(52\°)}\\ \\b=12*sin( 45\°)/sin( 52\°)\\ \\b=10.77\ units$

Step 3

Find the measure of side BC

Applying the law of sines

$\frac{a}{sin(A)}=\frac{c}{sin(C)}$

we have

a=BC

$c=12\ units$

$A=83\°$

$C=52\°$

substitute and solve for a

$\frac{a}{sin(83\°)}=\frac{12}{sin(52\°)}\\ \\a=12*sin(83\°)/sin(52\°)\\ \\a=15.11\ units$

Answer from: Quest

x=-2

infinite number of lines, so there is an infinite number of equations.

step-by-step explanation:

there are an infinite number of line that passes through the point (-2,5)

x=-2 is one line that passes through the point

y = 5 is another line that passes through the point

(these two lines are perpendicular to each other).

since the number of numbers are infinite and the slope is the ratio of two numbers,(change in y over change in x) we can have an infinite number of slopes. our only condition is that it passes through the point (-2,5)

A. write an equation of a line that passes through (-2, 5). b. how many different equations could y

Answer from: Quest

answer: the answer is 30 miles per hour.

ft/s -> mphmph = ft/s 1.466667mph = 44 divided by 1.466667mph = 30.00000

- r3ktforgood ☕

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1. consider the right triangle abc given below.
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