m∠B = 57.52°, m∠B = 70.8°, AB = 46.03 km and AC = 39.08 ft. This can be obtained using the Laws of cosine formula and Laws of sine formula.
Find the required angles and sides:
In a triangle ABC,
⇒ a² = b² + c² - 2bc cos A
⇒ b² = a² + c² - 2ac cos B
⇒ c² = a² + b² - 2ab cos C
where a, b and c are sides of a triangle and A, B and C are the angles of a triangle.
In a triangle ABC,
⇒ [tex]\frac{sinA}{a} =\frac{sinB}{b} =\frac{sinC}{c}[/tex]
where a, b and c are sides of a triangle and A, B and C are the angles of a triangle.
In the question we can use the laws of cosine formula and laws of sine formula,
5) Given that,
AB = c = 13 km
AC = b = 21 km
m∠A = 91°
By using Laws of cosine formula,
⇒ a² = b² + c² - 2bc cos A
a² = 21² + 13² - 2(21)(13) cos 91°
a² = 441 + 169 - 546 (-0.0174524064)
a² = 610 + 9.52901391 = 619.529014
⇒ a = 24.89 km
By using Laws of sine formula,
⇒ [tex]\frac{sinA}{a} =\frac{sinB}{b} =\frac{sinC}{c}[/tex]
sin 91°/24.89 = sin B/21
sin B = 0.999847695×21/24.89 = 0.843583833
⇒ m∠B = 57.52°
6) Given that,
AB = c = 11 cm
BC = a = 13 cm
AC = b = 14 cm
By using Laws of cosine formula,
⇒ b² = a² + c² - 2ac cos B
14² = 13² + 11² - 2(13)(11) cos B
196 = 169 + 121 - 286 cos B
196 = 290 - 286 cos B
cos B = 94/286
cos B = 0.328671329
⇒ m∠B = 70.8°
7) Given that,
AC = b = 24 km
BC = a = 26 km
m∠C = 134°
By using Laws of cosine formula,
⇒ c² = a² + b² - 2ab cos C
c² = 26² + 24² - 2(26)(24) cos 134°
c² = 676 + 576 - 1248 (-0.69465837)
c² = 1252 + 866.933646 = 2118.93365
⇒ AB = c = 46.03 km
8) Given that,
AB = c = 26 ft
BC = a = 21 ft
m∠B = 112°
By using Laws of cosine formula,
⇒ b² = a² + c² - 2ac cos B
b² = 21² + 26² - 2(21)(26) cos 112°
b² = 441 + 676 - 1096 (-0.374606593)
b² = 1117 + 410.568826 = 1527.56883
⇒ AC = b = 39.08 ft
Hence m∠B = 57.52°, m∠B = 70.8°, AB = 46.03 km and AC = 39.08 ft.
Learn more about Laws of cosine and Laws of sine here:
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