Question

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.

Answer 1

#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

Answer 2

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`

Explanation:

Part 1)

Part A

we know that

In the right triangle ABC

`sin(B)=(AC)/(AB)`

we have

`B=25\°`

`AC=b\ units`

`AB=25\ units`

Substitute and solve for b

`sin(25\°)=(b)/(25)`

`b=25*sin(25\°)=10.57\ units`

Part B

First way

we know that

In the right triangle ABC

`cos(B)=(BC)/(AB)`

we have

`B=25\°`

`BC=a\ units`

`AB=25\ units`

Substitute and solve for a

`cos(25\°)=(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)=(AC)/(BC)`

we have

`B=25\°`

`BC=a\ units`

`AC=b=10.57\ units`

substitute and solve for a

`tan(25\°)=(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

`(a)/(sin(A))=(c)/(sin(C))`

we have

`a=3\ units`

`c=1.35\ units`

`C=15\°`

substitute and solve for A

`(3)/(sin(A))=(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

so

`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

so

`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

`(b)/(sin(B))=(c)/(sin(C))`

we have

`b=AC`

`c=12\ units`

`B=45\°`

`C=52\°`

substitute and solve for b

`(b)/(sin(45\°))=(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

`(a)/(sin(A))=(c)/(sin(C))`

we have

`a=BC`

`c=12\ units`

`A=83\°`

`C=52\°`

substitute and solve for a

`(a)/(sin(83\°))=(12)/(sin(52\°))\\ \\a=12*sin(83\°)/sin(52\°)\\ \\a=15.11\ units`