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Given the triangle below, calculate the length of AB. 2. Solve triangle ABC, if a = 13, b = 11 and c = 16

Mathematics

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Answer

29.82 cm\text{29.82 cm}

A≈53.80∘,B≈43.10∘,C≈83.10∘A \approx 53.80^\circ, B \approx 43.10^\circ, C \approx 83.10^\circ

Step-by-step solution

7 steps

  1. Identify the given information for the triangle.

    We are given a triangle with sides BC=15BC = 15 cm, AC=25AC = 25 cm, and the included angle ∠C=93∘\angle C = 93^\circ. We need to find the length of side ABAB.

  2. Apply the Law of Cosines.

    The Law of Cosines states that for a triangle with sides a,b,ca, b, c and angles A,B,CA, B, C opposite those sides, c2=a2+b2−2abcos⁡(C)c^2 = a^2 + b^2 - 2ab \cos(C). In this case, a=BC=15a = BC = 15 cm, b=AC=25b = AC = 25 cm, and c=ABc = AB.

    AB2=BC2+AC2−2(BC)(AC)cos⁡(C)AB^2 = BC^2 + AC^2 - 2(BC)(AC) \cos(C) AB2=152+252−2(15)(25)cos⁡(93∘)AB^2 = 15^2 + 25^2 - 2(15)(25) \cos(93^\circ)

  3. Calculate the values.

    AB2=225+625−750cos⁡(93∘)AB^2 = 225 + 625 - 750 \cos(93^\circ) AB2=850−750(−0.052336)AB^2 = 850 - 750(-0.052336) AB2=850+39.252AB^2 = 850 + 39.252 AB2=889.252AB^2 = 889.252 AB=889.252AB = \sqrt{889.252} AB≈29.82 cmAB \approx 29.82 \text{ cm}

    The length of ABAB is 29.82 cm\boxed{\text{29.82 cm}}.

  4. Identify the given information for the triangle.

    We are given a triangle with sides a=13a = 13, b=11b = 11, and c=16c = 16. We need to find all unknown angles (A, B, C).

  5. Use the Law of Cosines to find angle A.

    The Law of Cosines states a2=b2+c2−2bccos⁡(A)a^2 = b^2 + c^2 - 2bc \cos(A).

    132=112+162−2(11)(16)cos⁡(A)13^2 = 11^2 + 16^2 - 2(11)(16) \cos(A) 169=121+256−352cos⁡(A)169 = 121 + 256 - 352 \cos(A) 169=377−352cos⁡(A)169 = 377 - 352 \cos(A) 352cos⁡(A)=377−169352 \cos(A) = 377 - 169 352cos⁡(A)=208352 \cos(A) = 208 cos⁡(A)=208352=1322\cos(A) = \frac{208}{352} = \frac{13}{22} A=arccos⁡(1322)≈53.80∘A = \arccos\left(\frac{13}{22}\right) \approx 53.80^\circ

  6. Use the Law of Cosines to find angle B.

    The Law of Cosines states b2=a2+c2−2accos⁡(B)b^2 = a^2 + c^2 - 2ac \cos(B).

    112=132+162−2(13)(16)cos⁡(B)11^2 = 13^2 + 16^2 - 2(13)(16) \cos(B) 121=169+256−416cos⁡(B)121 = 169 + 256 - 416 \cos(B) 121=425−416cos⁡(B)121 = 425 - 416 \cos(B) 416cos⁡(B)=425−121416 \cos(B) = 425 - 121 416cos⁡(B)=304416 \cos(B) = 304 cos⁡(B)=304416=1926\cos(B) = \frac{304}{416} = \frac{19}{26} B=arccos⁡(1926)≈43.10∘B = \arccos\left(\frac{19}{26}\right) \approx 43.10^\circ

  7. Find angle C using the sum of angles in a triangle.

    The sum of angles in a triangle is 180∘180^\circ.

    C=180∘−A−BC = 180^\circ - A - B C=180∘−53.80∘−43.10∘C = 180^\circ - 53.80^\circ - 43.10^\circ C=180∘−96.90∘C = 180^\circ - 96.90^\circ C=83.10∘C = 83.10^\circ

    The angles of the triangle are A≈53.80∘,B≈43.10∘,C≈83.10∘\boxed{A \approx 53.80^\circ, B \approx 43.10^\circ, C \approx 83.10^\circ}.

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