<span>The probability of the event that will not occur is 0.73.
</span>
#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
The rest of the question is the attached figure.
============================================
Δ AYW a right triangle at Y ⇒⇒⇒ ∴ WA² = AY² + YW²
And AY = YB ⇒⇒⇒ ∴ WA² = YB² + YW² → (1)
Δ BYW a right triangle at Y ⇒⇒⇒ ∴ WB² = BY² + YW² → (2)
From (1) , (2) ⇒⇒⇒ ∴ WA = WB →→ (3)
Δ CXW a right triangle at Y ⇒⇒⇒ ∴ WC² = CX² + XW²
And CX = XB ⇒⇒⇒ ∴ WC² = XB² + XW² → (4)
Δ BXW a right triangle at Y ⇒⇒⇒ ∴ WB² = XB² + XW² → (5)
From (4) , (5) ⇒⇒⇒ ∴ WC = WB →→ (6)
From (3) , (6)
WA = WB = WC
given ⇒⇒⇒ WA = 5x – 8 and WC = 3x + 2
∴ <span> 5x – 8 = 3x + 2</span>
Solve for x ⇒⇒⇒ ∴ x = 5
∴ WB = WA = WC = 3*5 + 2 = 17
The correct answer is option D. WB = 17
The one on the right would be greater
I don’t understand the question, can you restate it?
