(a) The triangle ABC has sides AB = 17m, BC = 12m and AC = 10m. Calculate the :
(i) largest angle of the triangle ; (ii) area of the triangle.
(b) From a point T on a horizontal ground, the angle of elevation of the top R of a tower RS, 38m high is 63°. Calculate, correct to the nearest metre, the distance between T and S.
(i) largest angle of the triangle ; (ii) area of the triangle.
(b) From a point T on a horizontal ground, the angle of elevation of the top R of a tower RS, 38m high is 63°. Calculate, correct to the nearest metre, the distance between T and S.
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Correct Answer: Option
Explanation:
(a) (i)
\(\cos C = \frac{a^{2} + b^{2} - c^{2}}{2ab}\)
\(\cos C = \frac{12^{2} + 10^{2} - 17^{2}}{2(12)(10)}\)
\(\cos C = \frac{-45}{240} = -0.1875\)
\(C = \cos^{-1} (-0.1875)\)
= \(100.81°\)
(ii) Area of \(\Delta ABC = \frac{1}{2} ab \sin C\)
= \(\frac{1}{2} \times 12 \times 10 \times \sin 100.81\)
= \(60 \times 0.9822\)
= \(58.936 m^{2}\)
(b)
\(\tan 63 = \frac{38}{x}\)
\(x = \frac{38}{\tan 63}\)
\(x = 19.362 m\)
\(\approxeq 19 m\)
(a) (i)
\(\cos C = \frac{a^{2} + b^{2} - c^{2}}{2ab}\)
\(\cos C = \frac{12^{2} + 10^{2} - 17^{2}}{2(12)(10)}\)
\(\cos C = \frac{-45}{240} = -0.1875\)
\(C = \cos^{-1} (-0.1875)\)
= \(100.81°\)
(ii) Area of \(\Delta ABC = \frac{1}{2} ab \sin C\)
= \(\frac{1}{2} \times 12 \times 10 \times \sin 100.81\)
= \(60 \times 0.9822\)
= \(58.936 m^{2}\)
(b)
\(\tan 63 = \frac{38}{x}\)
\(x = \frac{38}{\tan 63}\)
\(x = 19.362 m\)
\(\approxeq 19 m\)