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Inessa [10]
3 years ago
5

What is the area of a triangle that has a base of 15 1/4 in. and a height of 18in.

Mathematics
1 answer:
WARRIOR [948]3 years ago
8 0

Answer:

137.25 inch ²

Area of triangle = 1/2× base × height

= 1/2×15.25×18

= 15.25×9

=137.25 inch²

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Use the given information to find the exact value of the trigonometric function
eimsori [14]
\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}

Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

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Step-by-step explanation:

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