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natka813 [3]
3 years ago
6

How does the area of triangle RST compare to the area of triangle LMN?

Mathematics
2 answers:
RUDIKE [14]3 years ago
6 0

Answer:

The answer is the letter B or The area of △RST is equal to the area of △LMN.

Step-by-step explanation:


n200080 [17]3 years ago
6 0

Answer:

The correct option is 2. The area of △RST is equal to the area of △LMN.

Step-by-step explanation:

If three vertices of a triangle are given, then the area of triangle is

A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|

The vertices of triangle RST are R(5,5), S(2,1), T(1,3).

The area of △RST is

A=\frac{1}{2}|5(1-3)+2(3-5)+1(5-1)|

A=\frac{1}{2}|-10-4+4|

A=\frac{10}{2}

A=5

The area of △RST is 5 square unit.

The vertices of triangle LMN are L(0,-1), M(2,-4), N(-2,-3).

A=\frac{1}{2}|0(-4+3)+2(-3+1)-2(-1+4)|

A=\frac{1}{2}|-4-6|

A=\frac{10}{2}

A=5

The area of △LMN is 5 square unit.

The area of △RST is equal to the area of △LMN, therefore the correct option is 2.

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<h2>Answer:</h2>

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<h2>Step-by-step explanation:</h2>

From one of the trigonometric identities stated as follows;

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We can apply such identity to solve the given expression.

<em>Given:</em>

cos 28°​cos 62°​– sin 28°​sin 62°​

<em>Comparing the given expression with the right hand side of equation (i), we see that;</em>

A = 28°

B = 62°

<em>∴ Substitute these values into equation (i) to have;</em>

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<em />

<em>Solve the left hand side.</em>

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⇒ 0 = <em>cos28°cos62° - sin28°sin62°     (since cos 90° = 0)</em>

<em />

<em>Therefore, </em>

<em>cos28°cos62° - sin28°sin62° = 0</em>

<em />

<em />

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