Gce Model Questions Vol1 2020 Mathematics Question 36

Practice objective / multiple choice question 36 from the 2020 Gce Model Questions Vol1 Mathematics examination.

Gce Model Questions Vol1 2020 Mathematics Objective / Multiple Choice Hard Difficulty
Question 36 GCE_MODEL_QUESTIONS_VOL1 • 2020 • MATHEMATICS • objective

In triangle \(ABC\), angle \(A = 30^{\circ}\), side \(b = 12\) cm and side \(c = 15\) cm.\n(i) Find the length of side \(a\).\n(ii) Determine the radius \(R\) of the circumcircle of the triangle.\n(iii) Calculate the length of the altitude from \(A\) to side \(BC\). Give your answers to two decimal places where required.

Answer Options

A. a = \(\sqrt{369+180\sqrt3}\) \(\approx 20.12\) cm; R = \(\sqrt{369+180\sqrt3}\) \(\approx 20.12\) cm; altitude = \(\frac{90}{\sqrt{369+180\sqrt3}}\) \(\approx 4.47\) cm
B. a = \(\sqrt{369-180\sqrt3}\) \(\approx 7.56\) cm; R = \(\sqrt{369-180\sqrt3}\) \(\approx 7.56\) cm; altitude = \(\frac{180}{\sqrt{369-180\sqrt3}}\) \(\approx 23.80\) cm
Correct Answer C. a = \(\sqrt{369-180\sqrt3}\) \(\approx 7.56\) cm; R = \(\sqrt{369-180\sqrt3}\) \(\approx 7.56\) cm; altitude = \(\frac{90}{\sqrt{369-180\sqrt3}}\) \(\approx 11.90\) cm
D. a = \(\sqrt{369-180\sqrt3}\) \(\approx 7.56\) cm; R = \(\frac{a}{2\sin 30^{\circ}}\) \(\approx 15.12\) cm; altitude = \(\frac{90}{\sqrt{369-180\sqrt3}}\) \(\approx 11.90\) cm
Correct Answer
C
Correct Option:
a = \(\sqrt{369-180\sqrt3}\) \(\approx 7.56\) cm; R = \(\sqrt{369-180\sqrt3}\) \(\approx 7.56\) cm; altitude = \(\frac{90}{\sqrt{369-180\sqrt3}}\) \(\approx 11.90\) cm

About This Question

This is Gce Model Questions Vol1 2020 Mathematics Question 36. It is one of the objective questions from the 2020 Gce Model Questions Vol1 Mathematics examination.

Difficulty level: Hard .

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