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A vertical post is 20 m away from a point P on the ground. The angle of elevation of the top of the post from P is 30 degrees. Calculate the angle of elevation of the midpoint of the post from P.

Mathematics

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2 parts
a

1.5≤x<51.5 \le x < 5

b
2, 3, 4
2

Step-by-step solution

5 steps

Step 1: Determine the height of the post.
Let the height of the post be hh. Let P be the point on the ground, and T be the top of the post. The distance from P to the base of the post is 20 m. The angle of elevation of T from P is 30∘30^\circ.
Using trigonometry in the right-angled triangle formed:

tan⁡(θ)=oppositeadjacent\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} tan⁡(30∘)=h20\tan(30^\circ) = \frac{h}{20} h=20tan⁡(30∘)h = 20 \tan(30^\circ) h=20×13=203 mh = 20 \times \frac{1}{\sqrt{3}} = \frac{20}{\sqrt{3}} \text{ m}

Step 2: Determine the height of point S.
S is the midpoint of the post, so its height from the ground is half the total height of the post. Let the height of S be hSh_S.

hS=12h=12×203=103 mh_S = \frac{1}{2} h = \frac{1}{2} \times \frac{20}{\sqrt{3}} = \frac{10}{\sqrt{3}} \text{ m}

Step 3: Calculate the angle of elevation of S from P.
Let the angle of elevation of S from P be α\alpha.
Using trigonometry in the right-angled triangle formed by P, the base of the post, and S:

tan⁡(α)=hS20\tan(\alpha) = \frac{h_S}{20} tan⁡(α)=10/320\tan(\alpha) = \frac{10/\sqrt{3}}{20} tan⁡(α)=10203=123\tan(\alpha) = \frac{10}{20\sqrt{3}} = \frac{1}{2\sqrt{3}} α=arctan⁡(123)\alpha = \arctan\left(\frac{1}{2\sqrt{3}}\right) α≈arctan⁡(0.288675)\alpha \approx \arctan(0.288675) α≈16.102∘\alpha \approx 16.102^\circ

Rounding to 1 decimal place:

α≈16.1∘\alpha \approx 16.1^\circ

The angle of elevation of S from P is 16.1∘\boxed{\text{16.1}^\circ}.

Part a

Solve the inequalities.

The given combined inequality is x−5≤3x−8<2x−3x - 5 \le 3x - 8 < 2x - 3.
This can be split into two separate inequalities:

  1. x−5≤3x−8x - 5 \le 3x - 8
  2. 3x−8<2x−33x - 8 < 2x - 3

Solve the first inequality:

x−5≤3x−8x - 5 \le 3x - 8

Subtract xx from both sides:

−5≤2x−8-5 \le 2x - 8

Add 8 to both sides:

3≤2x3 \le 2x

Divide by 2:

32≤x\frac{3}{2} \le x x≥1.5x \ge 1.5

Solve the second inequality:

3x−8<2x−33x - 8 < 2x - 3

Subtract 2x2x from both sides:

x−8<−3x - 8 < -3

Add 8 to both sides:

x<5x < 5

Combine the solutions from both inequalities:

1.5≤x<51.5 \le x < 5

The solution to the inequalities is 1.5≤x<5\boxed{1.5 \le x < 5}.

Part b2, 3, 4

List the integral values of xx.

Integral values are integers. The integers that satisfy 1.5≤x<51.5 \le x < 5 are 2, 3, and 4.

The integral values of xx are 2,3,4\boxed{2, 3, 4}.

  1. Simplify the denominator.

    0.0608+0.0032=0.0640=0.0640.0608 + 0.0032 = 0.0640 = 0.064

  2. Simplify the numerator.

    0.4×1.280.4 \times 1.28

    We can write these as fractions:

    0.4=4100.4 = \frac{4}{10} 1.28=1281001.28 = \frac{128}{100} 0.4×1.28=410×128100=51210000.4 \times 1.28 = \frac{4}{10} \times \frac{128}{100} = \frac{512}{1000}

  3. Substitute the simplified numerator and denominator back into the expression.

    0.4×1.280.0608+0.00323=512/10000.0643\sqrt[3]{\frac{0.4 \times 1.28}{0.0608 + 0.0032}} = \sqrt[3]{\frac{512/1000}{0.064}}

    Convert 0.064 to a fraction:

    0.064=6410000.064 = \frac{64}{1000}

    Now substitute this back:

    512/100064/10003\sqrt[3]{\frac{512/1000}{64/1000}}

    When dividing by a fraction, multiply by its reciprocal:

    5121000×1000643\sqrt[3]{\frac{512}{1000} \times \frac{1000}{64}} 512643\sqrt[3]{\frac{512}{64}}

  4. Perform the division inside the cube root.

    512÷64=8512 \div 64 = 8

  5. Calculate the cube root.

    83=2\sqrt[3]{8} = 2

    The evaluated expression is 2\boxed{2}.

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