National Examination 2026 Math

Mathematics

ScanSolve AI Answer

Answer

22.00∘22.00^\circ

116.80∘116.80^\circ

41.19∘41.19^\circ

Step-by-step solution

3 steps

To find the three angles of the triangle, we will use the Law of Cosines. The given side lengths are a=8 ma = 8 \text{ m}, b=19 mb = 19 \text{ m}, and c=14 mc = 14 \text{ m}.

  1. Calculate angle A using the Law of Cosines.

    The formula for angle A is:

    cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}

    Substitute the given values:

    cos⁡A=192+142−822×19×14\cos A = \frac{19^2 + 14^2 - 8^2}{2 \times 19 \times 14} cos⁡A=361+196−64532\cos A = \frac{361 + 196 - 64}{532} cos⁡A=493532\cos A = \frac{493}{532} A=arccos⁡(493532)A = \arccos\left(\frac{493}{532}\right) A≈22.00∘A \approx \boxed{22.00^\circ}
  2. Calculate angle B using the Law of Cosines.

    The formula for angle B is:

    cos⁡B=a2+c2−b22ac\cos B = \frac{a^2 + c^2 - b^2}{2ac}

    Substitute the given values:

    cos⁡B=82+142−1922×8×14\cos B = \frac{8^2 + 14^2 - 19^2}{2 \times 8 \times 14} cos⁡B=64+196−361224\cos B = \frac{64 + 196 - 361}{224} cos⁡B=−101224\cos B = \frac{-101}{224} B=arccos⁡(−101224)B = \arccos\left(\frac{-101}{224}\right) B≈116.80∘B \approx \boxed{116.80^\circ}
  3. Calculate angle C using the Law of Cosines.

    The formula for angle C is:

    cos⁡C=a2+b2−c22ab\cos C = \frac{a^2 + b^2 - c^2}{2ab}

    Substitute the given values:

    cos⁡C=82+192−1422×8×19\cos C = \frac{8^2 + 19^2 - 14^2}{2 \times 8 \times 19} cos⁡C=64+361−196304\cos C = \frac{64 + 361 - 196}{304} cos⁡C=229304\cos C = \frac{229}{304} C=arccos⁡(229304)C = \arccos\left(\frac{229}{304}\right) C≈41.19∘C \approx \boxed{41.19^\circ}

    (Check: 22.00∘+116.80∘+41.19∘=179.99∘22.00^\circ + 116.80^\circ + 41.19^\circ = 179.99^\circ, which is approximately 180∘180^\circ).

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question