Step 1: Calculate angle ∠PRQ.
We use the Sine Rule:
sin(∠PRQ)PQ=sin(∠PQR)PR
Substitute the given values: PQ=41.6 m, PR=70 m, ∠PQR=105∘.
sin(∠PRQ)41.6=sin105∘70
Rearrange to solve for sin(∠PRQ):
sin(∠PRQ)=7041.6×sin105∘
Calculate sin105∘≈0.9659258:
sin(∠PRQ)=7041.6×0.9659258
sin(∠PRQ)=7040.20161328
sin(∠PRQ)≈0.57430876
Now, find ∠PRQ:
∠PRQ=arcsin(0.57430876)
∠PRQ≈35.05∘
Part (i): Calculate angle QPR.
The sum of angles in a triangle is 180∘.
∠QPR+∠PQR+∠PRQ=180∘
∠QPR=180∘−∠PQR−∠PRQ
Substitute the values:
∠QPR=180∘−105∘−35.05∘
∠QPR=39.95∘
Rounding to one decimal place:
∠QPR≈40.0∘
Part (ii): Calculate the area of triangle PQR.
The area of a triangle can be calculated using the formula:
Area=21absinC
Using sides PQ, PR and the included angle ∠QPR:
Area=21×PQ×PR×sin(∠QPR)
Substitute the values: PQ=41.6 m, PR=70 m, ∠QPR=39.95∘.
Area=21×41.6m×70m×sin(39.95∘)
Calculate sin(39.95∘)≈0.64207:
Area=21×41.6×70×0.64207
Area=20.8×70×0.64207
Area=1456×0.64207
Area≈934.78m2
Rounding to one decimal place:
Area \approx 934.8 \text{ m^2}
Part (iii): Calculate the shortest distance from Q to PR.
The shortest distance from a vertex to the opposite side is the altitude (height).
Let h be the shortest distance from Q to PR. The base is PR.
The area of a triangle can also be calculated as:
Area=21×base×height
Area=21×PR×h
We know the Area from Part (ii) and PR=70 m.
934.78m2=21×70m×h
934.78=35×h
Rearrange to solve for h:
h=35934.78
h≈26.708m
Rounding to one decimal place:
h≈26.7 m
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