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Mademuasel [1]
3 years ago
7

Consider the given right triangle. If angle B = 43 degrees and c = 26.7 meters, find a.

Mathematics
2 answers:
AVprozaik [17]3 years ago
5 0

Answer:

a ≈ 19.53 m

Step-by-step explanation:

The triangle is a right angle triangle . This means one angle is 90°. The picture below demonstrate a right angle triangle. If the angle B = 43° and side c = 26.7 meters the side a can be gotten below.

a = adjacent side

b = opposite side

c = hypotenuse

To find side a we use SOHCAHTOA principle

cos 43° = adjacent/hypotenuse

cos 43° = a/26.7

a = 26.7 × cos 43°

a = 26.7 ×  0.73135370161

a = 19.5271438332

a ≈ 19.53

Nady [450]3 years ago
3 0

Answer: Side a equals 19.5 metres

Step-by-step explanation: Consider the right angled triangle as shown in the picture attached. The triangle has been drawn with angle measuring 43 degrees, side c (line AB) measuring 26.7 m and side a (line CB) is yet unknown.

A right angled triangle can be solved if at least one side and an angle are available. In this question we shall apply the trigonometric ratios since we have one angle which shall be the reference angle (43°). Also we have an hypotenuse (the side facing the right angle) and an unknown side which is the adjacent (which lies between the right angle and the reference angle).

Cos B = Adjacent/Hypotenuse

Cos 43 = a/26.7

Cos 43 x 26.7 = a

0.7314 x 26.7 = a

19.52714 = a

a ≈ 19.5  (rounded to the nearest tenth)

Therefore the length of side a equals 19.5 metres.

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8 0
3 years ago
Read 2 more answers
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JulsSmile [24]

Answer:

20

Step-by-step explanation:

Our expression:

2 * (18 + - 8)

Plus then minus (+ -) is the same as just minus ( - )

Some examples:

1 + (+ 1) = 2

1 + (- 1) = 0

1 - (+ 1) = 0

1 - (- 1) = 2

Thus, 18 + - 8 is the same as 18 - 8

2 * (18 + - 8) =

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6 0
3 years ago
If
UkoKoshka [18]

Answer:

-8n + 9

Step-by-step explanation:

Given that,

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( - ) × ( - ) = ( + )

( + ) × ( - ) = ( - )

Let us solve now.

A - B

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Combine like terms

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Hope this helps you.

Let me know if you have any other questions :-)

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