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4 years ago

In right triangle ABC, ∠A = 90° and ∠C is 18° less than twice the measure of ∠B. What is the measure of ∠B?

Mathematics

1 answer:

spayn [35]4 years ago

6 0

Answer:

B = 36

Step-by-step explanation:

C = 2B -18 is the direct translation of the statement given in the question.

Since the sum of all angles of a triangle is 180.

A+B+C = 180

90 + B + (2B-18) = 180

solve for B

B = 36

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Step-by-step explanation:

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Read 2 more answers

Please help me find the area of shaded region and step by step

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Answer:

Part 1) $A=60\ ft^2$

Part 2) $A=80\ cm^2$

Part 3) $A=96\ m^2$

Part 4) $A=144\ cm^2$

Part 5) $A=9\ m^2$

Part 6) $A=(49\pi -33)\ in^2$

Step-by-step explanation:

Part 1) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior rectangle

The area of rectangle is equal to

$A=bh$

where

b is the base of rectangle

h is the height of rectangle

$A=(12)(7)-(8)(3)$

$A=84-24$

$A=60\ ft^2$

Part 2) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior square

The area of square is equal to

$A=b^2$

where

b is the length side of the square

$A=(12)(8)-(4^2)$

$A=96-16$

$A=80\ cm^2$

Part 3) we know that

The area of the shaded region is equal to the area of four rectangles plus the area of one square

$A=4(4)(5)+(4^2)$

$A=80+16$

$A=96\ m^2$

Part 4) we know that

The shaded region is equal to the area of the complete square minus the area of the interior square

$A=(15^2)-(9^2)$

$A=225-81$

$A=144\ cm^2$

Part 5) we know that

The area of the shaded region is equal to the area of triangle minus the area of rectangle

The area of triangle is equal to

$A=\frac{1}{2}(b)(h)$

where

b is the base of triangle

h is the height of triangle

$A=\frac{1}{2}(6)(7)-(6)(2)$

$A=21-12$

$A=9\ m^2$

Part 6) we know that

The area of the shaded region is equal to the area of the circle minus the area of rectangle

The area of the circle is equal to

$A=\pi r^{2}$

where

r is the radius of the circle

$A=\pi (7^2)-(3)(11)$

$A=(49\pi -33)\ in^2$

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