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Flura [38]
3 years ago
5

Triangle A B C is shown. Angle A C B is a right angle. The length of hypotenuse A B is 12 centimeters, the length of C B is 9.8

centimeters, and the length of A C is 6.9 centimeters.
Which expressions can be used to find m∠BAC? Select three options.

Mathematics
1 answer:
RSB [31]3 years ago
3 0

Answer: We can find angle BAC by using (1) SinA = 9.8/12

(2) CosA = 6.9/12

(3) TanA = 9.8/6.9

Step-by-step explanation: The question indicates that we have a right angled triangle ABC with the right angle at point C (that is, angle ACB is the right angle). Also the three sides have been labeled as AB equals 12 cm, CB equals 9.8 cm and AC equals 6.9 cm. Line AB has been identified as the hypotenuse and angle A (that is, BAC) is the reference angle. If angle A is the reference angle, then line CB facing angle A shall be the opposite side, while line AC is the adjacent (which lies between the reference angle and the right angle). Please see the attached diagram for details.

Having these details available, we can actually find angle A by using any of the three trigonometric ratios, since all three sides are given. Hence,

SinA = opposite/hypotenuse

SinA = 9.8/12

SinA = 0.8167

Checking with your calculator or table of values, A = 54.76 (approximately 55)

Also CosA = adjacent/hypotenuse

CosA = 6.9/12

CosA = 0.5750

Checking with your calculator or table of values A = 54.90 (approximately 55).

Finally TanA = opposite/adjacent

TanA = 9.8/6.9

TanA = 1.4203

Checking with your calculator or table of values A = 54.83 (approximately 55).

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Mariana [72]
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3 years ago
How many times does X squared minus 4X -12 cross the X axis
kolbaska11 [484]

We can solve this problem using discriminant.

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This means that it crosses over the axes 2 times.

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4 0
3 years ago
Differential cos^2x dy/dx =e^y-tanx​
gulaghasi [49]

Answer:

y=t−1+ce

−t

where t=tanx.

Given, cos

2

x

dx

dy

+y=tanx

⇒

dx

dy

+ysec

2

x=tanxsec

2

x ....(1)

Here P=sec

2

x⇒∫PdP=∫sec

2

xdx=tanx

∴I.F.=e

tanx

Multiplying (1) by I.F. we get

e

tanx

dx

dy

+e

tanx

ysec

2

x=e

tanx

tanxsec

2

x

Integrating both sides, we get

ye

tanx

=∫e

tanx

.tanxsec

2

xdx

Put tanx=t⇒sec

2

xdx=dt

∴ye

t

=∫te

t

dt=e

t

(t−1)+c

⇒y=t−1+ce

−t

where t=tanx

8 0
2 years ago
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