Please help me thank you

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# Question

# Answer

Hello, Sec 5 student here.

a) Apply Cosine Rule (c^{2 }= a^{2 }+ b^{2}-2abcos C)

- Since we know the angle and two sides, it’ll be:
- c
^{2}=16^{2}+15^{2}-2(16)(15)(cos 110) - To make it simpler, I’ll square root the whole equation, solve it and get:
- c≈25.4m //Keep in mind that before I’ve arrived to this, c=25.4001 in 5 significant figures (5 s.f.).

b) Apply Sine Rule [sin A/A = sin B/ B]

- We have our sin A, A length and B length. All we need is the value of sin B.
- Hence… sin
^{-1 }[(sin110/25.4001)(16)]=36.3° //For accuracy, I’ve used the 5 s.f. version of our part A answer.

c) For this, you’ll need to know the formula for volume of a Triangular Prism (0.5abh)

- We have our A and B, which will be our Base Area.
- However, we need to find height.
- For height, simply make a line perpendicular to line AC from angle ABC as that perpendicular bi-sector will serve as our height.
- Therefore… h(height)=[(15/sin90 due to right-angle)(sin 36.3) //Note that in Trigonometry, angles are required to have one decimal place unless it is an integer (whole number).
- Which will give us a height of 8.90m //8.8801m for 5 s.f.

Now that we have our height, we can solve part (c).

- Using the height, the volume of the figure will be:
- V (volume)= 0.5(25.4001)(24)(8.8801)
- Which will give us a volume of 2710m
^{3 }corrected to 3 s.f.

d) Ah, angle of elevation. I highly suggest drawing a separate figure for this sort of questions.

- Since it is an angle of elevation question, along with the pole (which I’ll name D
_{1}) is vertical, we can safely assume that the diagram will be a right-angled triangle. - Moreover, since we know the adjacent and opposite sides of the triangle, along with angle D
_{1}ED, all we need to know is the hypotenuse which can be solved by Pythagoras Theorem (c=√a^{2}+b^{2}). - Hence, c=√18
^{2}+15^{2} - Which will equal to 23.4m //23.4307 in 5 s.f.

Therefore…

- Angle of elevation is…
- sin
^{-1}[(sin90/23.407)(18)] - Which will equal to 50.2° corrected to one decimal place (1 d.p.)

e) You’ll have to find the Total Surface Area of the figure.

Let’s review what we have to calculate:

- Areas of… ΔABC+ΔFED +BEDC+AFEB+FDCA
- Fortunately, the two triangles are identical.
- Hence, ΔABC=ΔFED

To calculate the triangles, simply use 2(0.5bh) where b is the base and h is the height. //The reason why I added a 2 in front is to multiply the equation by 2 to account for the two identical triangles.

- Therefore, 2[0.5(25.4001)(8.8801)] would be 225.555… (this means that I’ve used the store function on my calculator.)

To calculate squares BEDC and AFEB simply use base*height.

- Which will be BEDC + AFEB…
- 15(24)+16(24) = 774m
^{2}

To calculate square FDCA, simply use base*height again.

- Which will be 25.4001(24)=609.6024m
^{2}

Now, we could simply amalgamate (combine) the values of each component of the figure, divide it by 450cm^{2 }and multiply the value acquired with the cost of the paint.

- Which will look like this:
- [(225.555… + 774 + 609.6024)/450] (30.25)
- Which will give you $106.15 corrected to 2 decimal places.

Therefore, painting the figure would cost $106.15.

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