Can I get some help here.

A gentleman is standing 10 feet from a streetlight. He is 5'6 tall and has a shadow of 24 feet. What is the height of the street

Goshia [24]

Answer:

Height of the streetlight ≈ 8 ft(nearest foot)

Step-by-step explanation:

The doc file displays the triangle formed from the illustration. x is the height of the street light. The distance from the gentle man to the street light is 10 ft. He has a height of 5.6 ft and the shadow formed on the ground is 24 ft long. The height of the street light can be calculated below.

The length of the tip of the shadow to the base of the street light is 34 ft. Similar triangle have equal ratio of their corresponding sides .

ab = 5.6 ft

The ratio of the base sides = 24/34

The ratio of the heights = 5.6/x

The two ratio are equal Therefore,

24/34 = 5.6/x

24x = 5.6 × 34

24x = 190.4

divide both side by 24

x = 190.4/24

x = 7.93333333333

x ≈ 8 ft

Height of the streetlight ≈ 8 ft(nearest foot)

Download docx

7 0

3 years ago

Easy geometric shapes question!

Evgen [1.6K]

Answer:

Cylinder

Step-by-step explanation:

The given geometric shape is of a cylinder.

8 0

3 years ago

Read 2 more answers

Identify the property that justifies step 1 when solving the equation

Ierofanga [76]

Answer:

Distributive Property

Step-by-step explanation:

Answer choice a is correct because you are distributing 2 to each term in the parentheses. When you multiply 2 to each factor in the parentheses, it simplifies to 2x-2, thus making the new equation 6x+2x-2=30

3 0

3 years ago

Show the work pleasee

Nataly [62]

Answer:

a

Step-by-step explanation:

6 0

3 years ago

State the number of possible triangles that can be formed using the given measurements.

romanna [79]

Answer: 39) 1 40) 2

41) 1 42) 0

<u>Step-by-step explanation:</u>

39) ∠A = ? ∠B = ? ∠C = 129°

a = ? b = 15 c = 45

Use Law of Sines to find ∠B:

$\dfrac{\sin B}{b}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin B}{15}=\dfrac{\sin 129}{45}\rightarrow \quad \angle B=15^o\quad or \quad \angle B=165^o$

If ∠B = 15°, then ∠A = 180° - (15° + 129°) = 36°

If ∠B = 165°, then ∠A = 180° - (165° + 129°) = -114°

Since ∠A cannot be negative then ∠B ≠ 165°

∠A = 36° ∠B = 15° ∠C = 129° is the only valid solution.

40) ∠A = 16° ∠B = ? ∠C = ?

a = 15 b = ? c = 19

Use Law of Sines to find ∠C:

$\dfrac{\sin A}{a}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin 16}{15}=\dfrac{\sin C}{19}\rightarrow \quad \angle C=20^o\quad or \quad \angle C=160^o$

If ∠C = 20°, then ∠B = 180° - (16° + 20°) = 144°

If ∠C = 160°, then ∠B = 180° - (16° + 160°) = 4°

Both result with ∠B as a positive number so both are valid solutions.

Solution 1: ∠A = 16° ∠B = 144° ∠C = 20°

Solution 2: ∠A = 16° ∠B = 4° ∠C = 160°

41) ∠A = ? ∠B = 75° ∠C = ?

a = 7 b = 30 c = ?

Use Law of Sines to find ∠A:

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{7}=\dfrac{\sin 75}{30}\rightarrow \quad \angle A=13^o\quad or \quad \angle A=167^o$

If ∠A = 13°, then ∠C = 180° - (13° + 75°) = 92°

If ∠A = 167°, then ∠C = 180° - (167° + 75°) = -62°

Since ∠C cannot be negative then ∠A ≠ 167°

∠A = 13° ∠B = 75° ∠C = 92° is the only valid solution.

42) ∠A = ? ∠B = 119° ∠C = ?

a = 34 b = 34 c = ?

Use Law of Sines to find ∠A:

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{34}=\dfrac{\sin 119}{34}\rightarrow \quad \angle A=61^o\quad or \quad \angle A=119^o$

If ∠A = 61°, then ∠C = 180° - (61° + 119°) = 0°

If ∠A = 119°, then ∠C = 180° - (119° + 119°) = -58°

Since ∠C cannot be zero or negative then ∠A ≠ 61° and ∠A ≠ 119°

There are no valid solutions.

6 0

3 years ago