O lvl E Maths: Trigonometry

In the diagram, ADC is a straight line. AB = 11 cm, BD = 10 cm and CD = 3 cm. Using as much of the information given below as is necessary, calculate
(i) sin ∡BAD,
(ii) BC²,
(iii) the area of triangle BCD

given that sin ∡ADB = 0.88, cos ∡ADB = -0.47, tan ∡ADB = -1.88

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Answer:

(i) Using sine rule,
sin ∡BAD / 10 = sin ∡ADB / 11
sin ∡BAD = 0.88 / 11 * 10
sin ∡BAD = 0.8


(ii) cos ∡BDC = - cos ∡ADB = 0.47

Using cosine rule,
BC² = BD² + CD² - 2 (BD)(CD)(cos ∡BDC)
BC² = 10² + 3² - 2(10)(3)(0.47)
BC² = 80.8


(iii) sin ∡BDC = sin ∡ADB = 0.88

Area of triangle BCD
= ½(BD)(CD) sin ∡BDC
= ½(10)(3) (0.88)
= 13.2 cm²

O lvl E Maths: Trigonometry

RGS P2 Prelim Q6a 1997

A church tower stands on horizontal ground and a point T at the top of the tower is 25 metres vertically above a point M of the base. The points A and B are on the same level as M. The point A is due south of the tower and the angle of elevation of T from A is 27°. The point B is due west of the tower and the angle of elevation of T from B is 35°.

(i) Calculate to one decimal place, the distances of AM, BM and BA respectively.

(ii) Calculate to the nearest degree, the bearing of B from A.

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Answer:

(i)


25/AM = tan 27°
AM = 25/tan 27° = 49.07 m



25/BM = tan 35°
AM = 25/tan 35° = 35.7 m


BA² = BM² + AM²
Thus, BA = √(35.7² + 49.07²)
BA = 60.7 m


(ii) angle MAB = tan^-1 35.7 / 49.07 = 36.04°
Hence, bearing of B from A = 360° - 36.04° = 324°

Note: Angle BMA is 90° because B is west of M and A is south of M

O lvl E Maths: Trigonometry

Question from http://www.sgforums.com/forums/2297/topics/331504

A and B are 2 houses, 40 m apart, on the same horizontal line as the foot F of a building TF. The angles of depression of A and B from top of building are 25 deg and 58 deg respectively. Find the possible heights of the building.

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Answer:

There are two different ways for the houses to be placed

Taking the difference of the ground distance,
h / tan 25° - h / tan 58° = 40
h (1 / tan 25° - 1 / tan 58°) = 40
1.5196 h = 40
h = 26.3 m

Taking the summation of the ground distance,
h / tan 25° + h / tan 58° = 40
h (1 / tan 25° + 1 / tan 58°) = 40
2.769 h = 40
h = 14.4 m

O lvl E Maths: Trigonometry

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Answer:

O level E Maths: Trigonometry

This is a standard question from http://www.sgforums.com/forums/2297/topics/330037

Given that tan x = 1/p and that x is not acute, p > 0, find

(i) sin(-x)

(ii) sin(90-x)

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Answer:

For tan x = 1/p, we can use the triangle method to find that the magnitude of sin x = 1 / √(1 +p²)

However, because x is not acute, and 1/p > 0, x lies in the 3rd quadrant, 180deg <>

So for that quadrant, sin x is negative. This gives us sin x = -1 / √(1 +p²)

Since sin (-x) = -sin x,
sin (-x) = 1 / √(1 +p²)

Note: It is now positive

sin (90 - x) = cos x
From the same triangle, find that the magnitude of cos x = p / √(1 +p²)

However, since x is in the 3rd quadrant,
cos x = -p / √(1 +p²)