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2 years ago

Find the measure of each angle in Triangle ABC. m∠a=(40x−21)°m∠b=(31−2x)°m∠c=(x+14)°

Mathematics

1 answer:

jarptica [38.1K]2 years ago

6 0

Answer:

D. m∠a=139°, m∠b=23°, m∠c=18°

Step-by-step explanation:

This is a multiple-choice question that can be answered simply by considering the reasonableness of the answers.

The expression for ∠a suggests it will be by far the largest of the angles, eliminating choices A and C.

No angle in a triangle is more than 180°, eliminating choice B.

The expression for ∠b suggests it will have a measure less than 31°. This confirms choice D as the correct one.

m∠a=139°, m∠b=23°, m∠c=18°

If you want to work the problem, you use the fact that the sum of angles is 180°

∠a +∠b +∠c = 180°

(40x -21)° +(31 -2x)° +(x +14)° = 180°

39x +24 = 180 . . . . . . . . . divide by °, collect terms

39x = 156 . . . . . . . . . . subtract 24

x = 4 . . . . . . . . . . . divide by 39

Then the measures of the angles are ...

∠a = (40×4 -21)° = 139°

∠b = (31 -2×4)° = 23°

∠c = (4 +14)° = 18°

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Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

$\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$

If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

$\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Hence:

$\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x$