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3 years ago

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The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is five times

the measure of the first angle. The third angle is 10 more than the second.

1 answer:

erastova [34]3 years ago

6 0

Answer:

first angle be 30°

second angle be 70°

third angle be 80°

Step-by-step explanation:

first angle be x

second angle be y

third angle be z

x+y+z=180

given, y+z = 5x; z=5x-y

also, z=y+10

5x-y = y+10

5x= 10+y+y = 10+2y

x= (10+2y)/5

x+y+z=180

(10+2y)/5 + y + y+10 = 180

(10+2y)/5 = 180 -2y -10

(10+2y)/5 = 170-2y

multiply both sides by 5

10+2y = 5(170-2y)

10+2y= 850-10y

2y+10y = 850-10

12y= 840

y= 840/12 = 70°

z=y+10 =70+10= 80°

5x= (10+2y) = 10+2(70) = 10+140= 150

x = 150/5 = 30°

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Answer:

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Step-by-step explanation:

Given

See attachment

From the attachment, we have:

$MN = 6$

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First, we need to calculate length LM,

Using Pythagoras theorem:

$LN^2 = MN^2 + LM^2$

$10^2 = 6^2 + LM^2$

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Collect Like Terms

$LM^2 = 100 - 36$

$LM^2 = 64$

$LM = 8$

Solving (a): $\sin L$

$\sin L = \frac{Opposite}{Hypotenuse}$

$\sin L = \frac{MN}{LN}$

Substitute values for MN and LN

$\sin L = \frac{6}{10}$

$\sin L = 0.60$

Solving (b): $tan\ N$

$tan\ N = \frac{Opposite}{Adjacent}$

$tan\ N = \frac{LM}{MN}$

Substitute values for LM and MN

$tan\ N = \frac{8}{6}$

$tan\ N = 1.33$

Solving (c): $\cos L$

$\cos L = \frac{Adjacent}{Hypotenuse}$

$\cos L = \frac{LM}{LN}$

Substitute values for LN and LM

$\cos L = \frac{8}{10}$

$\cos L = 0.80$

Solving (d): $\sin N$

$\sin N = \frac{Opposite}{Hypotenuse}$

$\sin N = \frac{LM}{LN}$

Substitute values for LM and LN

$\sin N = \frac{8}{10}$

$\sin N = 0.80$

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