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2 years ago

What is the area of triangle rst? 6 square units 9 square units 12 square units 18 square units

Mathematics

1 answer:

Butoxors [25]2 years ago

6 0

The area of the considered triangle RST is given by: Option A: 9 sq. units.

<h3>How to find the area of a triangle?</h3>

If we have:

Length of its base = b units
Its height = h units long,

Then we get:

Area of a triangle = $\dfrac{b \times h}{2} \: \rm unit^2$

The missing image is attached below.

If we take the base of the triangle as RS, then as the line UT is perpendicular to RS (as RS is parallel to x axis and UT is parallel to y axis) and touching from the base line RS to T, thus, it can be taken as height.

Thus, we have:

Area of RST = half of length of RS (base) times length of UT (height)

As visible from image, we have:

Length of RS = 6 blocks of unit length = 6 units

Similarly, length of UT = 3 units

Thus, we get:

Area of RST = $\dfrac{6 \times 3}{2} = 9 \: \rm unit^2$

Thus, the area of the considered triangle RST is given by: Option A: 9 sq. units.

Learn more about area of a triangle here:

brainly.com/question/14503701

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Answer:

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$ .

Thus:

$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

$=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }$

Let's do the first part first: (Recall how to divide fractions)

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For the second term:

$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

$\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}$

Note the numerator; it can be derived from the Pythagorean Identity:

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Thus, we can substitute the numerator:

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Everything simplifies to 1.

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