Aptitude - Height and Distance - Discussion

First, understand the problem from the given Diagram.

We have to find the distance of the CD.
So, first identify how to use 45° and 35°.
Only by using tan because we know AB value
We know that tanx = perpendicular/base.

tan 30° = AB/CA.
1/√3= 100/CA.
CA=100 x √3.
tan 45° = AB/AD.
1 = 100/AD,
AD = 100.
We know that CD = CA+CD.
= 100x√ 3+100,
= 100(√3+1),
= 100(1.73+1),
= 100(2.73),
= 273m.

From the figure shown above

sin30=(BA/BC); sin45=(BA/BD)

=> (1/2)=(100/BC); =>(0.7071)=(100/BD)

=>BC=200m; =>BD=141.42

by pythagorean theorem:

in a rt. angled triangle, the sum of squares of the two sides

(smaller sides) of a triangle is equal to the square of third

side(larger side).

in triangle BCA

(BC)^2 = (BA)^2 +(AC)^2

(200)^2=(100)^2+(AC)^2

=>AC=173.2m

in triangle BDA

(BD)^2 = (BA)^2 +(AD)^2

(141.42)^2=(100)^2+(AC)^2

=>AC=100m

Distance between two boats=AC+AD

=>173m+100m = 273m

Just try to understand.

We need to find AC and we know only AB.

BY P. Theorem, perpendicular/base, i.e.. AB/AC i.e., = tan.

1) We didn't know the value CB, if we take AB/CB no AC omit this.

2) AC/CB both value not known so no chance.

3) There is nothing called AC/AB mean base/perpendicular cancel this.

That's it try to read formula section you will understand.

According to My observation, when the angle of elevation is 45 degree then base and perpendicular must be equal which is 100 according to AD measurement if another elevation is less then from another which is 30 degree of AC then how is possible that distance/length of AC increase from another distance. Please anyone explain.

@All.

Know, how we get the 'tan'.
In the question is given we have to find the distance then that time we are not treating the angle of elevation so we have to take distance and hight then the formula becomes;

Tan=opp/adj.

Then AB/AC look at the figure it will be clear for you.

Section A of 5th standard has a class strength of 40 with an average height of 100.5 cm. Section B has a class.

Strength of 50 with an average height of 96 cm. What is the weighted average height of students in the 5th.

Standard? (Rounded to the nearest 0.5 cm).

Then, AB = 100 m, ACB = 30° and ADB = 45°.

AB = tan 30° = 1, AC = AB x 3 = 1003 m.
AC = 1003 m.
AB = tan 45° = 1, AD = AB = 100 m.
AD
CD = (AC + AD) = (1003 + 100) m.
= 100(3 + 1).
= (100 x 2.73) m.
= 273 m.

That Proved.

Short tricks
(1)30_60_90
1:√3:2.

So,
∆adb
a = 1×100.
= 100m.

b = √3×100.
= 173m.

d = 100 × 2
= 200m.

(2) 45_45_90
1:1:√2.

∆dcb
C = 1×100.
= 100m.

B = 1×100,
= 100m.

D = √2×100.
= 141m.

Total
ADC = 173 + 100,
= 273m.

@Kanishka Roy.

The one which is opposite to angle will be height.

And one which is adjacent to angle will be base.

Note that the one which is the base is not the longest side of triangle otherwise it would be the hypotenuse.

Hello friends. As best of my knowledge. Here we find the distance between two ships i.e. (AC+AD).

Here we know the height of the lighthouse (AB=100m). For the sake of we taken only (TAN = AB/AC). But not take. Sin&Cos.