Question

solve for the unknown parts of the triangle (i asked this question earlier but decided to up the pts hoping someone would answer T---T)

Answer 1

Answer: 1: ∠L = 53, LA = 23.15 ft, BA = 18.9 ft, 2: ∠S = 57.31, WS = 69.36 cm, SA = 58.37 cm, 3: AB = 19.33 cm, ∠B = 30.71 ,∠C = 99.29

Explanation:

For triangle 1, solving for the missing angle is quite simple since all internal angles of a triangle add up to 180, so 180 - 102 - 25 = 53. For triangles that aren't right-angle triangles, the first thing we want to do is add a line segment that bisects an angle and is perpendicular to one side. This will separate the triangle into two right-angle triangles and allow us to use Sin, Cos, and Tan to find missing angles and sides. In the case of traingle 1, I split line segment LA creating angle "C" (random letter). For the triangle LBC, I needed to find LC, so using Cosθ = Adjacent/Hypotenuse, I got Cos(53) = x/10 = 6.02. Before solving for BA and AC I need to find the side length of LB -> Tanθ = Opposite/Adjacent -> Tan(53) = x/6.02 -> x = 7.99. Now I can solve for BA and CA: BA -> Sin(25) = 7.99/x -> x = 18.9, CA -> Tan(25) = 7.99/x = 17.13. Now I have all my angles and sides besides LA which is just adding LC and AC -> 6.02 + 17.13 = 23.15.

Triangle 2, same process: ∠S = 180 - 79 - 43.69 = 57.31

SA = SB + BA -> WB: Sin(43.69) = x/84.5 = 58.37

AB: Tan(43.69) = 58.37/x = 61.1

BS: Tan(57.31) = 58.37/x = 37.45

WS: Sin(57.31) = 58.37/x = 69.36

AS: AB + BS = 98.55

Triangle 3, similar process, just involves finding an extra angle:

AB = AT + TB -> CT: Sin(50) = x/10 = 7.66

∠ACT: Cos^-1(7.66/10) = 40

AT: Cos(50) = x/10 = 6.43

∠BCT: Cos^-1(7.66/15) = 59.29

∠B: sin^-1(7.66/15) = 30.71

BT: Tan(30.71) = 7.66/x = 12.9

AB: AT + BT = 6.43 + 12.9 = 19.33

∠C = ∠BCT + ∠ACT = 40 + 59.29 = 99.29

I hope this helps! I am greatly sorry if something doesn't add up or make sense, and will fix it if need be just lmk! (Image is example of line segment)

Answer 2

1) ∠L = 53°

AL = 23.14 ft

AB = 18.90 ft

2) ∠S = 53.71°

SW = 69.35 cm

AS = 98.56 cm

3) ∠B = 30.71°

∠C = 99.29°

AB = 19.32 cm

Explanation:

To solve for the unknown parts of the triangles, we can use the Angle Sum Property of a Triangle and the Sine Rule.

The Angle Sum Property of a Triangle states that the interior angles of a triangle always sum to 180°.

`\boxed{\begin{minipage}{7.6 cm}\underline{Sine Rule} \\\\$(a)/(\sin A)=(b)/(\sin B)=(c)/(\sin C)$\\\\\\where:\\ \phantom{ww}$\bullet$ $A, B$ and $C$ are the angles. \\ \phantom{ww}$\bullet$ $a, b$ and $c$ are the sides opposite the angles.\\\end{minipage}}`

Question 1

Given values of triangle ABL:

• A = 25°
• B = 102°
• LN = 10 ft

As we have been given two angles, we can use the Angle Sum Property of a Triangle to find the measure of the third angle, ∠L.

`\angle A + \angle B + \angle L = 180^(\circ)`

`25^(\circ) + 102^(\circ) + \angle L = 180^(\circ)`

`127^(\circ) + \angle L = 180^(\circ)`

`\angle L = 53^(\circ)`

To find the lengths of sides AB and AL, use the Sine Rule.

`(LN)/(\sin A)=(AL)/(\sin B)=(AB)/(\sin L)`

`(10)/(\sin 25^(\circ))=(AL)/(\sin 102^(\circ))=(AB)/(\sin 53^(\circ))`

Therefore:

`AL=(10\sin 102^(\circ))/(\sin 25^(\circ))=23.1449440...=23.14\sf \; ft`

`AB=(10\sin 53^(\circ))/(\sin 25^(\circ))=18.8973260...=18.90\; \sf ft`

`\hrulefill`

Question 2

Given values of triangle ASW:

• A = 43.69°
• W = 79°
• AW = 84.5 cm

As we have been given two angles, we can use the Angle Sum Property of a Triangle to find the measure of the third angle, ∠S.

`\angle A + \angle S + \angle W = 180^(\circ)`

`43.69^(\circ) + \angle S+79^(\circ) = 180^(\circ)`

`122.69^(\circ) + \angle S= 180^(\circ)`

`\angle S = 57.31^(\circ)`

To find the lengths of sides AS and SW, use the Sine Rule.

`(SW)/(\sin A)=(AW)/(\sin S)=(AS)/(\sin W)`

`(SW)/(\sin 43.69^(\circ))=(84.5)/(\sin 57.31^(\circ))=(AS)/(\sin 79^(\circ))`

Therefore:

`SW=(84.5\sin 43.69^(\circ))/(\sin 57.31^(\circ))=69.3542652...=69.35\; \sf cm`

`AS=(84.5\sin 79^(\circ))/(\sin 57.31^(\circ))=98.5586958...=98.56\; \sf cm`

`\hrulefill`

Question 3

Given values of triangle ABC:

• A = 50°
• BC = 15 cm
• AC = 10 cm

To find angle B, use the Sine Rule.

`(BC)/(\sin A)=(AC)/(\sin B)=(AB)/(\sin C)`

`(15)/(\sin 50^(\circ))=(10)/(\sin B)=(AB)/(\sin C)`

`\sin B=(10\sin 50^(\circ))/(15)`

`B=\sin^(-1)\left((10\sin 50^(\circ))/(15)\right)=30.7102207...^(\circ)=30.71^(\circ)`

As we now have two angles, we can use the Angle Sum Property of a Triangle to find the measure of the third angle, ∠C.

`\angle A + \angle B + \angle C = 180^(\circ)`

`50^(\circ) + 30.71^(\circ) +\angle C= 180^(\circ)`

`80.71^(\circ) + \angle C= 180^(\circ)`

`\angle C = 99.29^(\circ)`

To find the length of the third side, AB, use the Sine Rule.

`(BC)/(\sin A)=(AC)/(\sin B)=(AB)/(\sin C)`

`(15)/(\sin 50^(\circ))=(10)/(\sin 30.71^(\circ))=(AB)/(\sin 99.29^(\circ))`

`AB=(15\sin 99.29^(\circ))/(\sin 50^(\circ))=19.3242...=19.32\;\sf cm`