pls pahelp po nito kailangan q na talaga ito ngayon

LEARNING TASK 2
Direction: Solve each of the following triangles. Round off your answers to the nearest hundredths. Show
your solution.

1. a = 11,

b = 14,

<C = 105°

2. a = 16,

b = 13

C = 20

» The given oblique triangle gives two sides and an included angle (SAS), solve for the third side then the following angles using the law of cosines.

Find side a:

[tex] \rm {c}^{2} = {a}^{2} + {b}^{2} - 2ab( \cos C) [/tex]

[tex] \rm {c}^{2} = 11^{2} + 14^{2} - 2(11)(14)( \cos 105 \degree) [/tex]

[tex] \rm {c}^{2} = 121 + 196 - 308( \cos 105 \degree) [/tex]

[tex] \rm {c}^{2} \approx 121 + 196 + 79.72[/tex]

[tex] \rm {c}^{2} \approx 396.72[/tex]

[tex] \rm \sqrt{{c}^{2} \: } \approx \sqrt{396.72 \: }[/tex]

[tex] \rm c \approx 19.92[/tex]

[tex] \therefore [/tex] Side c is approximately 19.92 units.

[tex] \rm [/tex]

Find ∠A:

[tex] \rm \cos A = \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} \\ [/tex]

[tex] \rm \cos A \approx \frac{14^{2} + (19.92)^{2} - 11^{2} }{2(14)(19.92)} \\ [/tex]

[tex] \rm \cos A \approx \frac{196 + 396.72 - 121 }{2(14)(19.92)} \\ [/tex]

[tex] \rm \cos A \approx \frac{ \: 471.72 \: }{557.76} \\ [/tex]

[tex] \rm \angle A \approx \cos^{ \text - 1} \bigg( \frac{ \: 471.72 \: }{557.76} \bigg) \\ [/tex]

[tex] \rm \angle A \approx 32.25 \degree[/tex]

[tex] \therefore [/tex] Angle A is approximately 32.25 degrees.

[tex] \rm [/tex]

Find ∠B

[tex] \rm \cos B = \frac{ a^{2} + {c}^{2} - b^{2} }{2ac} \\ [/tex]

[tex] \rm \cos B \approx \frac{11^{2} + (19.92)^{2} - 14^{2} }{2(11)(19.92)} \\ [/tex]

[tex] \rm \cos B \approx \frac{121 + 396.72 - 196 }{2(11)(19.92)} \\ [/tex]

[tex] \rm \cos B \approx \frac{ \: 321.72 \: }{438.24} \\ [/tex]

[tex] \rm \angle B \approx \cos^{ \text - 1} \bigg( \frac{ \: 321.72 \: }{438.24} \bigg) \\ [/tex]

[tex] \rm \angle B \approx 88.25 \degree[/tex]

[tex] \therefore[/tex] Angle B is approximately 88.25 degrees.

[tex]\red{••••••••••••••••••••••••••••••••••••••••••••••••••}[/tex]

Triangle 2:

» The given oblique triangle gives three sides (SSS), solve for the angles using the law of cosines.

Find ∠A:

[tex] \rm \cos A = \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} \\ [/tex]

[tex] \rm \cos A = \frac{13^{2} + 20^{2} - 16^{2} }{2(13)(20)} \\ [/tex]

[tex] \rm \cos A = \frac{169 + 400 - 256}{2(13)(20)} \\ [/tex]

[tex] \rm \cos A = \frac{ \:313 \: }{520} \\ [/tex]

[tex] \rm \angle A = \cos^{ \text - 1} \bigg(\frac{ \:313 \: }{520} \bigg) \\ [/tex]

[tex] \rm \angle A = 53 \degree [/tex]

[tex] \therefore [/tex] Angle A is 53 degrees.

[tex] \rm [/tex]

Find ∠B:

[tex] \rm \cos B = \frac{ a^{2} + {c}^{2} - b^{2} }{2ac} \\ [/tex]

[tex] \rm \cos B = \frac{16^{2} + 20^{2} - 13^{2} }{2(16)(20)} \\ [/tex]

[tex] \rm \cos B = \frac{256 + 400 - 169}{2(16)(20)} \\ [/tex]

[tex] \rm \cos B = \frac{ \: 487 \: }{640} \\ [/tex]

[tex] \rm \angle B = \cos^{ \text - 1} \bigg(\frac{ \: 487 \: }{640} \bigg)\\ [/tex]

[tex] \rm \angle B \approx 40.45 \degree[/tex]

[tex] \therefore [/tex] Angle B is approximately 40.45 degrees.

[tex] \rm [/tex]

Find ∠C:

[tex] \rm \angle A + \angle B + \angle C = 180 \degree [/tex]

[tex] \rm 53 \degree + 40.45 \degree + \angle C \approx 180 \degree [/tex]

[tex] \rm 93.45 \degree + \angle C \approx 180 \degree [/tex]

[tex] \rm \angle C \approx 180 \degree - 93.45 \degree[/tex]

[tex] \rm \angle C \approx 86.55\degree[/tex]

[tex] \therefore [/tex] Angle C is approximately 86.55 degrees.

[tex]\red{••••••••••••••••••••••••••••••••••••••••••••••••••}[/tex]

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