Hello, Sec 5 student here.
a) Apply Cosine Rule (c2 = a2 + b2-2abcos C)
- Since we know the angle and two sides, it’ll be:
- c2=162+152-2(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 2710m3 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 D1) 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 D1ED, all we need to know is the hypotenuse which can be solved by Pythagoras Theorem (c=√a2+b2).
- Hence, c=√182+152
- Which will equal to 23.4m //23.4307 in 5 s.f.
- Angle of elevation is…
- 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) = 774m2
To calculate square FDCA, simply use base*height again.
- Which will be 25.4001(24)=609.6024m2
Now, we could simply amalgamate (combine) the values of each component of the figure, divide it by 450cm2 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.