8.3: Solution of Right Triangles(Heights and distances):

In this section we learn methods of finding values of remaining sides and angles when we are given:

1. Two sides of the triangle.

2. One side and one interior (standard) angle of the triangle.

8.3 Problem 1: Find angle x in the below mentioned figure.

Solution:

= tan 60 = Perp./Base

= DC/AD = 30/AD

AD = 30/

sin x = Perp./Hypot.

= AD/AB = (30/)/10

= 3/ =

Thus sin x =

8.3 Problem 2: In the figure given below, a rocket is fired vertically upwards from its launching pad P. It first rises 40km vertically and then travels 40km at 60⁰ to the vertical.

PA represents the first stage of the journey and AB the second. C is a point vertically below B on the horizontal level as P.

Calculate:

(!) The height of the rocket from the ground when it is at point B

(!!) The horizontal distance of point C from P.

Solution:

Since PA is parallel to CB

CD = 40km and

ABD = 60⁰and hence BAD =30⁰(Alternate angles)

Cos 60 = Base /Hyp. = BD/AB = BD/40

We know cos 60 = 1/2

1/2 = BD/40

Thus BD =20

sin 60 = Perp./Hyp. = AD/40

Since sin 60 = /2

We have

/2 = AD/40

AD = 20 = 20*1.732 = 34.64 km

BC = BD+CD= 20+40 = 60km

8.3 Problem 3: A ladder is placed against a vertical tower. If the ladder makes an angle of 30⁰with the ground and reaches up to a height of 15m of the tower,

find the length of the ladder.

Solution:

It is given that BD=15 and DAB = 30⁰

Sin 30 = sin DAB = Perp./Hyp. = BD/AD = 15/AD

We know sin 30 = 1/2

1/2 = 15/AD

AD = 30m

8.3 Problem 4: A kite is attached to a 100m long string.Find the greatest height reached by the kite, when the string makes an angle of 60⁰ with the level ground

Solution:

We know

Sin 60 = Perp./Hyp. = Max height/100

We also know sin 60 = /2

Max height/100 = /2

Thus

Max height = 100/2 = 50 = 50*1.732 = 86.6 m

8.3 Problem 5: If tan x = 5/12, tan y = 3/4 and AB = 48m find the length of CD

Solution:

tan x = Perp/Base = DC/AC

tan y = Perp/Base = DC/BC

tanx/tan y =(DC/AC)/ (DC/BC ) = (DC/AC) *(BC/DC)= BC/AC

By substituting given values for tanx and tan y we get

tan x/tan y=(5/12)/(3/4) = (5/12)*(4/3) = 5/9

BC/AC = 5/9 I.e. 9BC = 5AC

Since AC=AB+BC, 5AC = 5(AB+BC) = 5AB+5BC

9BC = 5AB+5BC I.e. 4BC= 5AB = 5*48 = 240 BC = 60m

tan y = Perp/Base = DC/BC = DC/60

But it is given that tan y = 3/4

Hence 3/4 = DC/60 DC = (3/4)*60 = 45M

8.3 Problem 6: The perimeter of a rhombus is 96cm and obtuse angle of it is 120⁰. Find the length of its diagonals

Solution:

Since in a rhombus all sides are equal :PQ = 96/4 = 24cm Let PQR = 120⁰

We also know that in rhombus diagonals bisect each other perpendicularly and diagonals bisect the angle at vertex.

Hence POR is a right angled triangle and PQO = 1/2(PQR) = 60⁰

Sin 60 = Perp./Hyp. = PO/PQ = PO/24

But sin 60 = /2 PO/24 = /2

i.e. PO = 12 = 12*1.732 = 20.784

PR = 2PO = 2*20.784 = 41.568cm

cos 60 = Base/Hyp. = OQ/24

But cos 60 = 1/2 OQ/24 = 1/2

i.e. QO = 24 /2 =12

QS = 2QO = 2*12 = 24cm

8.3 Problem 7: As observed from the top of 150M tall light house, the angles of depression of two ships approaching it are 30⁰ and 45⁰

If one ship is behind the other, find the distance between the two ships.

(The angle of depression is the angle made by the imaginary horizontal line to the ground from the top of light house to the ship, as seen by observer sitting in the light house).

Solution:

In the adjoining figure CO can be imagined to be light house of height 150m.

B and A are the position of ships.

XOA is the angle of depression of the ship A which is given to be 30⁰

XOB is the angle of depression of the ship B which is given to be 45⁰

Since OX is parallel to the ground

It follows that OAC = 30⁰ OBC = 45⁰ OC is the opposite side =150

BC/150= Cot 45 = 1 (Cot 45 = 1) BC =150

AC/150= Cot 30 = (Cot 30 = ) AC =150

AB(Distance between ships) = AC-BC = 150 -150 = 150(-1) = 109.8(Approximate)

8.3 Summary of learning

No	Points studied
1	Finding of angles and sides

Their Name:
Their Email:
Your Email:
Your Name:
Your message to your friend: