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LenaWriter [7]
2 years ago
10

Find the area of triangle

Mathematics
1 answer:
liraira [26]2 years ago
8 0

So we see <u>one large triangle, that can be split into two triangles split by the thin dotted line</u>:

  • the<em> First triangle</em> --> has a hypotenuse of 26 and a leg of 11
  • the <em>Second triangle</em> --> has a hypotenuse of 42

Let us find the <u>other leg of the </u><u><em>First triangle</em></u> which also happens to be the <u>height of the larger triangle</u>, using the Pythagorean theorem

     a^2+b^2=c^2\\a^2 + 11^2 = 26^2\\a^2 = 26^2-11^2=555\\a=23.56

Thus the <u>height of the entire triangle</u> is 23.56 which also happens to be <u>one of the legs of the </u><u><em>Second triangle</em></u>

<em>      </em>a^2+b^2=c^2\\a^2+(23.56)^2=42^2\\a^2 = 1764-555=1209\\a = 34.77

Having found the other leg of the <em>Second Triangle</em>, we can find the base of the larger triangle is:

     -->  34.77 + 11 = 45.77

The height as found in the first initial step is 23.56 thus the area is

   Area = \frac{1}{2}*b*h=\frac{1}{2}*23.56*45.77=539.17

Hope that helps!

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Answer:

Step-by-step explanation:

Let's take a look at the given angle 135°

The sketch of the angle which corresponds to -\dfrac{3\pi}{4} unit circle and can be seen in the attached image below;

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Trigonometric ratio           related ratio on coordinate axes

sin θ                                     \dfrac{y}{1}

cos θ                                    \dfrac{x}{1}

tan θ                                     \dfrac{y}{x}

csc θ                                     \dfrac{1}{y}

sec θ                                     \dfrac{1}{x}

cot θ                                      \dfrac{x}{y}

From the sketch of the image attached below;

The six trigonometric ratio for 135° can be expressed as follows:

sin (-\dfrac{3\pi}{4})= \dfrac{y}{1}

sin (-\dfrac{3\pi}{4})=- \dfrac{\sqrt{2}}{2}

cos (-\dfrac{3\pi}{4})= \dfrac{x}{1}

cos (-\dfrac{3\pi}{4})= -\dfrac{\sqrt{2}}{2}

tan (-\dfrac{3\pi}{4})= \dfrac{y}{x}

tan (-\dfrac{3\pi}{4})= \dfrac{-\dfrac{\sqrt{2}}{2}}{-\dfrac{\sqrt{2}}{2}}

tan (-\dfrac{3\pi}{4})= -\dfrac{\sqrt{2}}{2}} \times {-\dfrac{2}{\sqrt{2}}

tan (-\dfrac{3\pi}{4})= 1

csc (-\dfrac{3\pi}{4})= \dfrac{1}{y} \\ \\ csc (-\dfrac{3\pi}{4})=\dfrac{1}{-\dfrac{\sqrt{2}}{2}} \\ \\ csc=1 \times -\dfrac{2}{\sqrt{2}} \\ \\csc =-\sqrt{2}

sec (-\dfrac{3 \pi}{4})=\dfrac{1}{x} \\ \\  sec = \dfrac{1}{(-\dfrac{\sqrt{2}}{2})} \\ \\ sec = 1 \times -\dfrac{2}{\sqrt{2}}  \\ \\  sec = - \sqrt{2}

cot(-\dfrac{3 \pi}{4}) = \dfrac{x}{y} \\ \\ cot(-\dfrac{3 \pi}{4}) = \dfrac{-\dfrac{\sqrt{2}}{2} }{-\dfrac{\sqrt{2}}{2}} \\ \\  cot(-\dfrac{3 \pi}{4})=  -\dfrac{\sqrt{2}}{2} } \times {-\dfrac{2}{\sqrt{2}}} \\ \\  cot (-\dfrac{3 \pi}{4}) = 1

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Answer:

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