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Serga [27]
3 years ago
10

What additional information would prove each pair of triangles congruent by the HL Theorem?

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
3 0

Answer:

Please don't report me!

I don't know if this will help, but here is the definition of the HL theorem.

The HL Theorem states that if the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the two triangles are congruent.

Step-by-step explanation:

Additional link to help! ;)

https://tutors.com/math-tutors/geometry-help/hl-theorem

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Show with work please.
kolbaska11 [484]

Answer:

$\csc \left(\theta-\frac{\pi }{2}\right)=0.73$

Step-by-step explanation:

The identity you will use is:

$\csc \left(x\right)=\frac{1}{\sin \left(x\right)}$

So,

$\csc \left(\theta-\frac{\pi }{2}\right)$

$\csc \left(\theta-\frac{\pi }{2}\right)=\frac{1}{\sin \left(-\frac{\pi }{2}+\theta\right)}$

Now, using the difference of sin

Note: state that \text{sin}(\alpha\pm \beta)=\text{sin}(\alpha) \text{cos}(\beta) \pm \text{cos}(\alpha) \text{sin}(\beta)

$\csc \left(\theta-\frac{\pi }{2}\right)=\frac{1}{-\cos \left(\theta\right)\sin \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)\sin \left(\theta\right)}$

Solving the difference of sin:

$-\cos \left(\theta\right)\sin \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)\sin \left(\theta\right)$

-\cos \left(\theta\right) \cdot 1+0\cdot \sin \left(\theta\right)

-\text{cos} \left(\theta\right)

Then,

$\csc \left(\theta-\frac{\pi }{2}\right)=-\frac{1}{\cos \left(\theta\right)}$

Once

\text{sec}(-\theta)=\text{sec}(\theta)

And, \text{sec}(\theta)=-0.73

$-\frac{1}{\cos \left(\theta\right)}=-\text{sec}(\theta)$

$-\frac{1}{\cos \left(\theta\right)}=-(-0.73)$

$-\frac{1}{\cos \left(\theta\right)}=0.73$

Therefore,

$\csc \left(\theta-\frac{\pi }{2}\right)=0.73$

3 0
3 years ago
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