Question

Sum number 6(a) and 7(c).
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Answer 1

Correction in (Q1):

In the figure, not drawn to scale, TF is a tower. The elevation of T from A is x°, where tan x = 2/5 and AF = 200m. The elevation of T from B, where AB = 80m, is y°. Calculate:

(i) The height of the tower TF.

(ii) The angle y, correct to the nearest degree.

Solution:

According to the question:

$\Longrightarrow \rm tanx = \dfrac{2}{5}$

We know that:

$\Longrightarrow \rm tanx = \dfrac{Side \ opposite \ to \ x}{Side \ adjacent \ to \ x}$

$\Longrightarrow \rm tanx = \dfrac{TF}{AF}$

Substitute tanx = 2/5 above.

$\Longrightarrow \rm \dfrac{2}{5} = \dfrac{TF}{AF}$

$\Longrightarrow \rm \dfrac{2}{5} = \dfrac{TF}{200}$

$\Longrightarrow \rm \dfrac{2 \times 200}{5} = TF$

$\Longrightarrow \rm 2 \times 40 = TF$

$\Longrightarrow \rm{TF = 80 \ meters.}$

∴ (i) The height of the tower is 80 meters.

In ΔTBF:

$\Longrightarrow \rm tany = \dfrac{Side \ opposite \ to \ y}{Side \ adjacent \ to \ y}$

$\Longrightarrow \rm tany = \dfrac{TF}{BF}$

$\Longrightarrow \rm tany = \dfrac{80}{AF - AB}$

$\Longrightarrow \rm tany = \dfrac{80}{200 - 80}$

$\Longrightarrow \rm tany = \dfrac{80}{120}$

$\Longrightarrow \rm tany = \dfrac{8}{12}$

$\Longrightarrow \rm tany = \dfrac{2}{3}$

$\Longrightarrow \rm y = tan^{-1} \times \dfrac{2}{3} \ degrees.$

           or

$\Longrightarrow \rm y \approx 34^{\circ}$

∴ The angle y, correct to the nearest degree is 34°.

_____________________

(Q2) 'A' can do a piece of work in x days and B can do it in (x + 16) days. If both start working together, they can do it in 15 days. Find 'x'.

Solution:

$\Longrightarrow \sf Work \ done \ by \ 'A' = \dfrac{1}{Time \ taken \ in \ days}$

$\Longrightarrow \sf Work \ done \ by \ 'A' = \dfrac{1}{x}$

$\Longrightarrow \sf Work \ done \ by \ 'B' = \dfrac{1}{Time \ taken \ in \ days}$

$\Longrightarrow \sf Work \ done \ by \ 'A' = \dfrac{1}{x + 16}$

If both of them work together, they can finish the work in 15 days.

$\Longrightarrow \sf Work \ done \ by \ 'A' + Work \ done \ by \ 'B' = \dfrac{1}{No: \ of \ days \ taken}$

$\Longrightarrow \sf \ \dfrac{1}{x} + \dfrac{1}{x + 16} = \dfrac{1}{15}$

Take LCM:

$\Longrightarrow \sf \ \dfrac{x + 16 + x}{x(x + 16)} = \dfrac{1}{15}$

$\Longrightarrow \sf \ \dfrac{2x + 16}{x^2 + 16x} = \dfrac{1}{15}$

Cross multiply:

$\Longrightarrow \sf 15(2x + 16) = x^2 + 16x$

$\Longrightarrow \sf 30x + 240 = x^2 + 16x$

$\Longrightarrow \sf x^2 + 16x - 30x - 240 = 0$

Split the middle term:

$\Longrightarrow \sf x^2 - 14x - 240 = 0$

Sum ---: -14

Product ---: -240

Split ---: 10 × -24

$\Longrightarrow \sf x^2 + 10x - 24x - 240 = 0$

$\Longrightarrow \sf x(x + 10) - 24(x + 10) = 0$

$\Longrightarrow \sf (x - 24) (x + 10) = 0$

Compare both the roots/solutions with zero.

Case I

$\Longrightarrow \sf x -24 = 0$

$\Longrightarrow \sf x = 24$

Therefore, x = 24.

Case II

$\Longrightarrow \sf x + 10 = 0$

$\Longrightarrow \sf x = -10$

This is an invalid case as time cannot be negative. therefore x ≠ 10.

Hence the value of 'x' is 24.