The angles and lengths of each of the given triangles are;
5) m∠B = 57.52°
6) B = 70.81°
7) AB = 55.43 Km
8) AC = 39.06 ft
<h3>How to use cosine rule?</h3>
The cosine rule is expressed as;
c = √[a² + b² - 2ab(cos C)]
5) Using cosine rule;
BC = √[21² + 13² - 2(21*13)(cos 91)]
BC = 24.89
Using sine rule, we can find angle B as;
21/sin m∠B = 24.89/sin 91
sin m∠B = (21 * sin 91)/24.89
sin m∠B = 0.8436
m∠B = sin⁻¹0.8436
m∠B = 57.52°
6) Using cosine rule;
14² = 11² + 13² - 2(11*13)(cos B)]
196 = 121 + 169 - 286(cos B)
cos B = (121 + 169 - 196)/286
cos B = 0.3287
B = cos⁻¹0.3287
B = 70.81°
7) Using cosine rule;
AB = √[24² + 36² - 2(24*36)(cos 134)]
AB = 55.43 Km
8) Using cosine rule;
AC = √[21² + 26² - 2(21*26)(cos 112)]
AC = 39.06 ft
Read more about cosine rule at; brainly.com/question/4372174
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A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a load of 8.00 kg is hung on its end. Find the stress, the strain, and the Young’s modulus for the material of the wire.
Solution:
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a lo
main
prev Statement of a problem № 2618 next
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a load of 8.00 kg is hung on its end. Find the stress, the strain, and the Young’s modulus for the material of the wire.
Solution:
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a lo
main
prev Statement of a problem № 2618 next
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a load of 8.00 kg is hung on its end. Find the stress, the strain, and the Young’s modulus for the material of the wire.
Solution:
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a lo
main
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Besides just using the math in everyday life, it is also used in newspapers.
Answer: 
Step-by-step explanation:
We are given the equation 
. We want to isolate V.
[multiply both sides by h]
[divide both sides by 3]
Now we know .
