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

Step-by-step 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

Answer:

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

Step-by-step 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°.

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

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.

$\dfrac{LN}{\sin A}=\dfrac{AL}{\sin B}=\dfrac{AB}{\sin L}$

$\dfrac{10}{\sin 25^{\circ}}=\dfrac{AL}{\sin 102^{\circ}}=\dfrac{AB}{\sin 53^{\circ}}$

Therefore:

$AL=\dfrac{10\sin 102^{\circ}}{\sin 25^{\circ}}=23.1449440...=23.14\sf \; ft$

$AB=\dfrac{10\sin 53^{\circ}}{\sin 25^{\circ}}=18.8973260...=18.90\; \sf ft$

$\hrulefill$

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.

$\dfrac{SW}{\sin A}=\dfrac{AW}{\sin S}=\dfrac{AS}{\sin W}$

$\dfrac{SW}{\sin 43.69^{\circ}}=\dfrac{84.5}{\sin 57.31^{\circ}}=\dfrac{AS}{\sin 79^{\circ}}$

Therefore:

$SW=\dfrac{84.5\sin 43.69^{\circ}}{\sin 57.31^{\circ}}=69.3542652...=69.35\; \sf cm$

$AS=\dfrac{84.5\sin 79^{\circ}}{\sin 57.31^{\circ}}=98.5586958...=98.56\; \sf cm$

$\hrulefill$

Question 3

Given values of triangle ABC:

• A = 50°
• BC = 15 cm
• AC = 10 cm

To find angle B, use the Sine Rule.

$\dfrac{BC}{\sin A}=\dfrac{AC}{\sin B}=\dfrac{AB}{\sin C}$

$\dfrac{15}{\sin 50^{\circ}}=\dfrac{10}{\sin B}=\dfrac{AB}{\sin C}$

$\sin B=\dfrac{10\sin 50^{\circ}}{15}$

$B=\sin^{-1}\left(\dfrac{10\sin 50^{\circ}}{15}\right)=30.7102207...^{\circ}=30.71^{\circ}$

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.

$\dfrac{BC}{\sin A}=\dfrac{AC}{\sin B}=\dfrac{AB}{\sin C}$

$\dfrac{15}{\sin 50^{\circ}}=\dfrac{10}{\sin 30.71^{\circ}}=\dfrac{AB}{\sin 99.29^{\circ}}$

$AB=\dfrac{15\sin 99.29^{\circ}}{\sin 50^{\circ}}=19.3242...=19.32\;\sf cm$