Quiz – 4

Adding the two equations,
tan ∅ = (m+n)2(m+n)2
Subtracting the same,
sin ∅ = (m−n)2(m−n)2
Since, there are no available direct formula for relation between sin∅ ∅tan∅ but we know that
cosec² ∅ – cos² ∅ = 1

∴ ((2(m−n)2(m−n))² - ((2(m+n)2(m+n))² = 1
=> 4((m+n)2−(m−n)2))((m2−n2))24((m+n)2−(m−n)2))((m2−n2))2 = 1
=>((m²-n²))² = 4(4mn)
=> m² - n² = 4√mn

The question is "To find the value of m²= n²"

Hence, the answer is 4√mn

the height of the building with the height of the student!

Let BC be the height of the tower and DC be the height of the student.

In rt. ∆ ABC
AB = BC cot 45^°
AB = 100 m ……………(i)

In rt. ∆ ABD
AB = BD cot 60^°
AB = (BC + CD) cot 60^°
AB = (10 + CD) * 1√31√3……………(ii)

Equating (i) and (ii)
(10 + CD) * 1√31√3 = 100
(10 + CD)= 100√3
CD = 100√3 – 100
= 10(1.732 – 1) = 100 * 0.732 = 73.2 m

The question is "To find the height of the student"

Hence, the answer is 73.2 m

Cos x – Sin x = √2 Sin x

=> Cos x = Sin x + √2 Sin x
=> Cos x = Sin x + √2 Sin x
=> Sin x = Cosx(√2+1)Cosx(√2+1) * Cos x
=> Sin x = (√2−1)(√2−1)(√2−1)(√2−1) * 1(√2+1)1(√2+1) * Cos x
=> Sin x = (√2−1)((√2)2−(1)2)(√2−1)((√2)2−(1)2) * Cos x
=> Sin x = (√2 - 1) Cos x
=> Sin x = √2 Cos x – Cos x
=> Sin x + Cos x = √2 Cos x

The question is "To find the value of Cos x + Sin x"

Hence, the answer is √2 Cos x

the broken tree to the opposite building can help!

Let the broken portion of tree AA’ be x. Hence A’C = A’G = x

From the figure, total height of the tree = x + y + 6
Consider triangle A’BG, tan 45° = A′BBGA′BBG
1 = y+6BGy+6BG
Or BG = y + 6

Consider triangle A’C’C, tan 30° = A′BC′CA′BC′C
1√31√3 = yBG+4yBG+4
1√31√3 = yy+6+4yy+6+4
y√3 = y + 10
or y = 10√3−110√3−1
= 10√3−110√3−1 * √3+1√3+1√3+1√3+1
= 10(√3+1)210(√3+1)2
Therefore y = 5(√3 + 1)
Take sin 30° to find x
sin 30° = A′C′A′CA′C′A′C
1212 = yxyx
or x = 2y
Height of the tree = x + y = 6
= 2 * 5(√3 + 1) + 5(√3 + 1) + 6
= 10√3 + 10 + 5√3 + 5 + 6
= 15√3 + 21 meters

The question is "To find the total height of the tree in meters before it broke"

Hence, the answer is 15√3 + 21

the building and the man!

Let height of the flag pole PQ be x
Height of building QR = 7√3 m
Height of man AB = √3
Therefore QQ’ = 7√3 – √3 = 6√3

From diagram, tan y° = QQ′AQ′QQ′AQ′
Or tan y° = 6√3√1356√3√135 ------- (1)
Also, tan (90 – y)° = PQ’/AQ’
Or tan (90 – y)° = (x+QQ′)√135(x+QQ′)√135
But tan (90 – y)° = cot y°
Therefore cot y = x+6√3√135x+6√3√135
or tan y = √135x+6√3√135x+6√3 --------(2)

Equating (1) and (2), we get
6√3√1356√3√135 = √135x+6√3√135x+6√3
6√3x + 108 = 135
6√3x = 27
x = 92√392√3 or x = 1.5√3

The question is "To find the height of the flag pole."

Hence, the answer is 1.5√3