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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 places b Find the length of side a to two decimal places in three diffe class=
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 diffe class=

Respuesta :

#1) 
A) b = 10.57
B) a = 22.66; the different methods are shown below.
#2)
A) Let a = the side opposite the 15° angle; a = 1.35.
Let B = the angle opposite the side marked 4; m∠B = 50.07°.
Let C = the angle opposite the side marked 3; m∠C = 114.93°.
B) b = 10.77
m∠A = 83°
a = 15.11

Explanation
#1)
A) We know that the sine ratio is opposite/hypotenuse.  The side opposite the 25° angle is b, and the hypotenuse is 25:
sin 25 = b/25

Multiply both sides by 25:
25*sin 25 = (b/25)*25
25*sin 25 = b
10.57 = b

B) The first way we can find a is using the Pythagorean theorem.  In Part A above, we found the length of b, the other leg of the triangle, and we know the measure of the hypotenuse:
a²+(10.57)² = 25²
a²+111.7249 = 625

Subtract 111.7249 from both sides:
a²+111.7249 - 111.7249 = 625 - 111.7249
a² = 513.2751

Take the square root of both sides:
√a² = √513.2751
a = 22.66

The second way is using the cosine ratio, adjacent/hypotenuse.  Side a is adjacent to the 25° angle, and the hypotenuse is 25:
cos 25 = a/25

Multiply both sides by 25:
25*cos 25 = (a/25)*25
25*cos 25 = a
22.66 = a

The third way is using the other angle.  First, find the measure of angle A by subtracting the other two angles from 180:
m∠A = 180-(90+25) = 180-115 = 65°

Side a is opposite ∠A; opposite/hypotenuse is the sine ratio:
a/25 = sin 65

Multiply both sides by 25:
(a/25)*25 = 25*sin 65
a = 25*sin 65
a = 22.66

#2)
A) Let side a be the one across from the 15° angle.  This would make the 15° angle ∠A.  We will define b as the side marked 4 and c as the side marked 3.  We will use the law of cosines:
a² = b²+c²-2bc cos A
a² = 4²+3²-2(4)(3)cos 15
a² = 16+9-24cos 15
a² = 25-24cos 15
a² = 1.82

Take the square root of both sides:
√a² = √1.82
a = 1.35

Use the law of sines to find m∠B:
sin A/a = sin B/b
sin 15/1.35 = sin B/4

Cross multiply:
4*sin 15 = 1.35*sin B

Divide both sides by 1.35:
(4*sin 15)/1.35 = (1.35*sin B)/1.35
(4*sin 15)/1.35 = sin B

Take the inverse sine of both sides:
sin⁻¹((4*sin 15)/1.35) = sin⁻¹(sin B)
50.07 = B

Subtract both known angles from 180 to find m∠C:
180-(15+50.07) = 180-65.07 = 114.93°

B)  Use the law of sines to find side b:
sin C/c = sin B/b
sin 52/12 = sin 45/b

Cross multiply:
b*sin 52 = 12*sin 45

Divide both sides by sin 52:
(b*sin 52)/(sin 52) = (12*sin 45)/(sin 52)
b = 10.77

Find m∠A by subtracting both known angles from 180:
180-(52+45) = 180-97 = 83°

Use the law of sines to find side a:
sin C/c = sin A/a
sin 52/12 = sin 83/a

Cross multiply:
a*sin 52 = 12*sin 83

Divide both sides by sin 52:
(a*sin 52)/(sin 52) = (12*sin 83)/(sin 52)
a = 15.11
calculista

Answer:

Part 1)

Part a) [tex]b=10.57\ units[/tex]

Part b) [tex]a=22.66\ units[/tex] (three different ways in the procedure)

Part 2)

First triangle (triangle a)

Part a) [tex]c=1.35\ units[/tex]  

Part b) [tex]A=35.11\°[/tex]

Part c) [tex]B=129.89\°[/tex]

Second triangle  (triangle b)

Part a) [tex]A=83\°[/tex]      

Part b) [tex]AC=10.77\ units[/tex]

Part c) [tex]BC=15.11\ units[/tex]  

Step-by-step explanation:

Part 1)

Part A

we know that

In the right triangle ABC

[tex]sin(B)=\frac{AC}{AB}[/tex]  

we have

[tex]B=25\°[/tex]

[tex]AC=b\ units[/tex]

[tex]AB=25\ units[/tex]

Substitute and solve for b

[tex]sin(25\°)=\frac{b}{25}[/tex]

[tex]b=25*sin(25\°)=10.57\ units[/tex]

Part B

First way

we know that

In the right triangle ABC

[tex]cos(B)=\frac{BC}{AB}[/tex]

we have

[tex]B=25\°[/tex]

[tex]BC=a\ units[/tex]

[tex]AB=25\ units[/tex]

Substitute and solve for a

[tex]cos(25\°)=\frac{a}{25}[/tex]

[tex]a=25*cos(25\°)=22.66\ units[/tex]

Second way

Applying the Pythagoras theorem

[tex]c^{2}=a^{2} +b^{2}[/tex]

we have

[tex]c=25\ units[/tex]

[tex]b=10.57\ units[/tex]

substitute and solve for a

[tex]25^{2}=a^{2} +10.57^{2}[/tex]

[tex]a^{2}=25^{2}-10.57^{2}[/tex]

[tex]a=22.66\ units[[/tex]

Third way

we know that

In the right triangle ABC

[tex]tan(B)=\frac{AC}{BC}[/tex]

we have

[tex]B=25\°[/tex]

[tex]BC=a\ units[/tex]

[tex]AC=b=10.57\ units[/tex]

substitute and solve for a

[tex]tan(25\°)=\frac{10.57}{a}\\ \\a=10.57/ tan(25\°)\\ \\a=22.66\ units[/tex]

Part 2)

triangle a

we have

[tex]C=15\°[/tex]

[tex]a=3\ units[/tex]  

[tex]b=4\ units[/tex]  

Step 1

Find the measure of length side c

Applying the law of cosines

[tex]c^{2}=a^{2}+b^{2}-2(a)(b)cos(C)[/tex]  

substitute

[tex]c^{2}=3^{2}+4^{2}-2(3)(4)cos(15\°)[/tex]    

[tex]c^{2}=25-24cos(15\°)[/tex]  

[tex]c=1.35\ units[/tex]    

Step 2

Find the measure of angle A

Applying the law of sines

[tex]\frac{a}{sin(A)}=\frac{c}{sin(C)}[/tex]

we have

[tex]a=3\ units[/tex]

[tex]c=1.35\ units[/tex]

[tex]C=15\°[/tex]

substitute and solve for A

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

Step 3

Find the measure of angle B      

Remember that the sum of the internal angles of a triangle must be equal to [tex]180[/tex] degrees

so

[tex]A+B+C=180\°[/tex]

we have

[tex]C=15\°[/tex]

[tex]A=35.11\°[/tex]

substitute

[tex]B=180\°-35.11\°-15\°=129.89\°[/tex]      

triangle b

we have

[tex]C=52\°[/tex]

[tex]B=45\°[/tex]

[tex]c=12\ units[/tex]  


Step 1  

Find the measure of angle A      

Remember that the sum of the internal angles of a triangle must be equal to [tex]180[/tex] degrees

so

[tex]A+B+C=180\°[/tex]

we have

[tex]C=52\°[/tex]

[tex]B=45\°[/tex]

substitute

[tex]A=180\°-52\°-45\°=83\°[/tex]    

Step 2

Find the measure of side AC

Applying the law of sines

[tex]\frac{b}{sin(B)}=\frac{c}{sin(C)}[/tex]

we have

[tex]b=AC[/tex]

[tex]c=12\ units[/tex]

[tex]B=45\°[/tex]

[tex]C=52\°[/tex]

substitute and solve for b

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

Step 3

Find the measure of side BC

Applying the law of sines

[tex]\frac{a}{sin(A)}=\frac{c}{sin(C)}[/tex]

we have

[tex]a=BC[/tex]

[tex]c=12\ units[/tex]

[tex]A=83\°[/tex]

[tex]C=52\°[/tex]

substitute and solve for a

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


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