3 years ago

If the area of this triangle is 15 square inches. What is the length of the base?

Mathematics

1 answer:

solong [7]3 years ago

4 0

Umm I believe 3?? Because 5x3=15

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Part 1) $sin(\theta)=\frac{2}{3}$

Par 2) $cos(\theta)=-\frac{\sqrt{5}}{3}$

Part 3) $tan(\theta)=-\frac{2\sqrt{5}}{5}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we have

$csc(\theta)=\frac{3}{2}$

Remember that

$csc(\theta)=\frac{1}{sin(\theta)}$

therefore

$sin(\theta)=\frac{2}{3}$

step 2

Find the $cos(\theta)$

we know that

$sin^{2}(\theta) +cos^{2}(\theta)=1$

we have

$sin(\theta)=\frac{2}{3}$

substitute

$(\frac{2}{3})^{2} +cos^{2}(\theta)=1$

$\frac{4}{9} +cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{4}{9}$

$cos^{2}(\theta)=\frac{5}{9}$

square root both sides

$cos(\theta)=\pm\frac{\sqrt{5}}{3}$

we have that

$cos(\theta) < 0$ ---> given problem

$cos(\theta)=-\frac{\sqrt{5}}{3}$

step 3

Find the $tan(\theta)$

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=\frac{2}{3}$

$cos(\theta)=-\frac{\sqrt{5}}{3}$

substitute

$tan(\theta)=\frac{2}{3}:-\frac{\sqrt{5}}{3}=-\frac{2}{\sqrt{5}}$

Simplify

$tan(\theta)=-\frac{2\sqrt{5}}{5}$

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3 years ago