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3 years ago

In a right triangle, Sin (30 + x) º = Cos (3x) º. What is the value of x?

Mathematics

1 answer:

pogonyaev3 years ago

5 0

Answer:

$x=15^\circ$

Step-by-step explanation:

In a right triangle:

$\sin \theta =cos(90^\circ-\theta) $ (Complementary angles)$

Therefore, given:

$Sin (30 + x) \º = Cos (3x) \º\\30+x+3x=90^\circ\\30+4x=90^\circ\\4x=90^\circ-30^\circ\\4x=60^\circ\\$Divide both sides by 4\\x=15^\circ$

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An infinite number of ratios are equivalent to 3:8. How

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Answer:

See below, please!

Step-by-step explanation:

There are infinite number of ratios equivalent to 3:8 because both numbers can be multiplied by a common number or divided by a common number.

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3 years ago

The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.

Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m + 40 m + c = 120 m

61 m + c = 120 m

c = 120 m - 61 m

c = 59 m

↓ ↓

A → → → → → → → → → → → C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

$a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0$

Angle B is calculated as follows;

$Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0$

Angle C is calculated as follows;

$Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0$

Therefore, the largest angle of the field is 149⁰.

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