## Angles and Plane Figures Questions

1.  The sum of angles of a triangle is given by the expression (2a+b)0 while that of a quadrilateral is given by . Calculate the values of a and b     (4 mks)

2.  The figure below represents a quadrilateral ABCD. Triangle ABX is an equilateral triangle. If , find with   (2 mks)

3.  Wanjiku is standing at a point P, 160m south of a hill H on a level ground. From point P she observes the angle of elevation of the top of the hill to be 670

(a) Calculate the height of the hill  (3 mks)

(b) After walking 420m due east to the point Q, Wanjiku proceeds to point R due east of Q, where the angle of elevation of the top of the hill is 350. Calculate the angle of elevation of the top of the hill from Q     (3 mks)

(c) Calculate the distance from P to R    (4 mks)

4.  In the triangle XYZ below, find the angle ZXY. (3mks)

5.  The exterior angle of a regular polygon is equal to one-third of the interior angle. Calculate the number of sides of the polygon and give its name. (4 mks)

6.  In the figure below, lines AB and LM are parallel.

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Find the values of the angles marked x, y and z  (3 mks)

7.  From points A and B on a level ground the angles of elevation to the top of the building are 240 and 380 respectively. If the distance between A and B is 47m and that of B from the foot of the building is X;

(a) Form an expression for the height of the building

(b) Calculate the height of the building

(c) Find the difference in the distance between the top of the building and points A and B

8.  The angle of elevation of the top of the tower from the foot of a building is 63.510. The angle of depression of the top of the building from the op of the tower is 18.430. The building and the tower are 30m apart. Find

1. The height of the tower   (1mk)
2. The height of the building (2mks)

9.  The exterior angle of a regular polygon is an eighth of the interior angle. How many sides does the regular polygon have?  (3 marks)

10.  The sides of a parallelogram are 4cm by 5cm and its area is 12cm2. Calculate its angles.  (3 marks)

11.  From a point 20m away on a level ground the angle of elevation to the lower window line is 270 and the angle of elevation to the top line of the window is 320. Calculate the height of the window.

(3 marks)

12.  A regular polygon has its exterior angle 180, and one of its sides 16cm. calculate its area.

(to 2 d.p) (3mks)

13.  The angle of depression of a point A on the ground from the top of a post is 180 and that of  another point B on the same line as A nearer to the foot of the post is 250. If A and B are 70m  apart,

(a) Draw a sketch to represent positions of A and B. (2mks)

(i) The height of the post from the ground level (ans 1 d.p) (6mks)

(ii) The distance of point A from the foot of the post. (2mks)

14.  The figure below shows an irregular polygon PQRSTUVW.

Calculate the sum of all the interior angles in the figure below.

15.  The angles of elevation from two points A and B to the top of a storey building are 480 and 570 respectively. If AB = 50m and the point A and B are opposite each other; Calculate;

a) the distance of point A to the building  (2 mks)

b) the height of the building (2 mks)

## Angles and Plane figures Answers

 1 M1 M1  M1A1 formation of the equationsattempt to solve 2 B1B1 3 h = 160 × tan 670 = 376.94m     Tan ө = M1     M1 A1         M1     M1A1      M1 M1       M1 A1 4. aA2 = b2 + c2 – 2bc cos A42 = 32 + 62 – 2 x 3 x 6 cos -29 = -36cos -29 = cos -3636.340=  M1 M1 A1 Substitution Attempt to simplify 03 5. The polygon is an octagon M1  M1 A13 6. B1B1B13

9.  Let the ex < be x0 ALT

In < 8x0  n
= No. of sides

n – 2 180 = 8 360 M1M1

x + 8x = 180…………..  M1   n n

x = 20 n = 18 sides A1

No of sides = 360 M1

20

= 18 sides A1

3

10.

5 h

B

4

Area = 5 x sin  = 12  M1

 = 36.870 A1

B = 143.130 A1

3

11.50 270

Window x

h

Tan 270 = h

20  M1

h = 10.19m

Tan 320 = x/20  M1

x = 12.50m

Window height = 2.31m A1

3

 12. 3600 = 180 nn = 3600 = 20 sides 180Area = ( ½ x 16 x 16/2 tan 810) x 20= (8 x8 x 6.3138) x 20= 8081.66cm2 B1 M1 A1 03 13. (a)Sketch(b) i)h = tan250  h = x tan 25xh = tan180  h = tan 18(x + 70)x+70equating the two equationsx tan 250 = x tan 180 + 70 tan 180x(tan 25 tan 180) = 70 tan 180x = 70 tan 180 tan250 – tan180x = 22.744 = 160.8 0.1414h = 160.8tan 25 = 75m(c) Distance of A to the front of post= x + 70= 160.8 + 70= 230.8m B2    M1 M1  M1 M1 M1 A1 B1  M1 A1 10 14. {2(8) – 2} x 90  14 x 90  12600 M1  A12 15. M1  A1  M1  A1 04

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