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BlackZzzverrR [31]
2 years ago
5

What is the length of the side of a right triangle that has a side length of 12ft and a hypotenuse that measures 15ft

Mathematics
2 answers:
icang [17]2 years ago
6 0

Once again, In order to find the length of the hypotenuse, we need to use Pythagorean's Theorm.

This theorm states that a^2+b^2=c^2

a and b are sides, and c is the hypotenuse

12^2 + b^2 = 15^2\\144+b^2 = 225\\b^2 = 81\\\sqrt{b^2} = \sqrt{81}\\b = 9


The other side is 9

muminat2 years ago
4 0

Answer: 9 ft


Step-by-step explanation:

 1. To solve this exercise you must apply the Pythagorean Theorem, which is:

a=\sqrt{b^{2}+c^{2}}

Where a is the hypotenuse, and b and c are the other sides of the triangle.

2. Then, when you  solve for one of the sides and substitute the values given in the problem into the formula shown above, you obtain that the length of  the side of the rigth triangle is:

 b=\sqrt{(15ft)^{2}-(12ft)^{2}}

 b=9ft



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A running circuit is in the shape of a triangle with lengths of 6km, 6.5km and 7km. What are the sizes of the angles (in minutes
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A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to 180^{o}. The <em>measures</em> of the internal <u>angles</u> of the <u>triangle</u> given in the question are A = 52.6^{o}, B = 59.4^{o}, and C = 68^{o}.

A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to 180^{o}.

Considering the given question, let the <u>sides</u> of the triangle be: a = 6 km, b = 6.5 km, and c = 7 km.

Apply the <em>Cosine rule</em> to have:

c^{2} = a^{2} + b^{2} - 2ab Cos C

So that;

7^{2} = 6^{2} + (6.5)^{2} - 2(6 * 6.5) Cos C

49 = 36 + 42.25 - 78Cos C

78 Cos C = 78.25 - 49

               = 29.25

Cos C = \frac{29.25}{78}

         = 0.375

C = Cos^{-1} 0.375

   = 67.9757

C = 68^{o}

Apply the <em>Sine rule</em> to determine the <u>value</u> of B,

\frac{b}{Sin B} = \frac{c}{Sin C}

\frac{6.5}{Sin B} = \frac{7}{Sin 68}

SIn B = \frac{6.5 *Sin 68}{7}

         = 0.861

B = Sin^{-1} 0.861

   = 59.43

B = 59.4^{o}

Thus to determine the value of A, we have;

A + B + C = 180^{o}

A + 59.4^{o} + 68^{o} = 180^{o}

A = 180^{o} - 127.4

  = 52.6

A = 52.6^{o}

Therefore the <u>sizes</u> of the <em>internal angles</em> of the triangle are: A = 52.6^{o}, B = 59.4^{o}, and C = 68^{o}.

For more clarifications on applications of the Sine and Cosine rules, visit: brainly.com/question/14660814

#SPJ1

8 0
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