Question

4a. A boy walks 5km due North and 4km due East
i. Find the bearing of his current position from the starting point.
ii. How far is the boy now from the starting point. 6 marks
b. The angle of elevation of the sun is 45 0 . A tree has a shadow 12 m long.
Find the height of the tree.

Answer 1

Answer:

a) i) North-East

ii) sq. root of 41

b) sq. root of 72/11

Step-by-step explanation:

a)

i) The boy is in North-East.

ii) Using pythagorus theorem on the triangle this formed,

(5)^(2) + (4)^(2) = x^(2) where x is the boy's distance from the starting point.

25 + 16 = x^(2)

41 = x^(2)

x = sq. root of 41.

b.

Forming a triangle from this we find x = 45 deg.

Opposite = 12 m

We have to find Adjacent side.

Sin(x) = Opp./Adj.

Sin(45) = 12/y, where y is a variable for the length of tree.

sq. root of 22 = 12/y, where we put in the value of sin(45)

Squaring both sides,

22 = 144/(y^(2))

y^(2) = 144/22

y^(2) = 72/11

y = sq. root of 72/11

Answer 2

4 a.

$\largw\underline{\underline{\sf{\color{orange}{GIVEN:-}}}}$

A boy walks 5km due North and 4km due East.

$\largw\underline{\underline{\sf{\color{orange}{TO\ FIND:-}}}}$

i. The bearing of his current position from the starting point.

ii. How far is the boy now from the starting point.

$\largw\underline{\underline{\sf{\color{orange}{SOLUTION:-}}}}$

i. For the bearing we need the measure of the angle CAB.

Now,

$\sf{tan\ A=\dfrac{p}{b} }$

$\sf{\to tan\ A=\dfrac{BC}{AC}}$

$\sf{\to tan\ A =\dfrac{5}{4}}$

$\sf{\to A=tan^-^1\bigg(\dfrac{5}{4}\bigg)}$

$\sf{\to \angle CAB=tan^-^1 \bigg(\dfrac{5}{4} \bigg)}$

___________________

ii. For distance of the boy from the starting point,

• AB = Base, b = 3 km
• BC = Perpendicular, p = 4 km
• AC = Hypotenuse, h = √(3² + 4²) = √25 = 5 km

The distance from starting point = AB = 5 km

____________________________

4 b.

$\largw\underline{\underline{\sf{\color{orange}{GIVEN:-}}}}$

• The angle of elevation of the sun is 45°.
• A tree has a shadow 12 m long.
• AB is the tree.

$\largw\underline{\underline{\sf{\color{orange}{TO\ FIND:-}}}}$

The height of the tree (AB).

$\largw\underline{\underline{\sf{\color{orange}{SOLUTION:-}}}}$

We know,

$\sf{tan\ C=\dfrac{p}{b}}$

$\sf{\to tan\ 45^\circ=\dfrac{AB}{BC}}$

$\sf{\to 1=\dfrac{AB}{12}}$

$\sf{\to AB=12\ cm}$

Therefore, the height of the tree is 12 cm.