Question

A3.6 ladder leans on a wall to reach the height of 7.2 m on the wall . If the ladder makes an angle of measure 30 with the ground ,  the length of the ladder is​

Answer 1

Laddernleans Q7. Soham painted the outside of the cabinet of measure 1.5 m x 3 m x 2.5 m. How much surface area
did he cover if he painted all except the bottom of the cabinet?

Answer 2

Correct Question:-

A ladder leans on a wall to reach the height of 7.2 on the wall. If thd ladder makes and angle of measure 30° with the ground, the length of the ladder is _________.

Given:-

• Height of the wall till the ladder reach on leaning = 7.2 m
• Angle of elevation from the ground = 30°

To Find:-

• The length of the ladder.

Note:-

• Refer to the attachment for figure.

Solution:-

In ∆ABC,

∠B is right - angled.

∠C = 30°

AB = 7.2

We know,

SinC = $\sf{\dfrac{Perpendicular}{Hypotenuse} = \dfrac{AB}{AC}}$

Hence,

$\sf{Sin30^\circ = \dfrac{AB}{AC}}$

From trigonometric table we have:-

• Sin30° = $\sf{\dfrac{1}{2}}$

Putting the value,

= $\sf{\dfrac{1}{2} = \dfrac{7.2}{AB}}$

On cross - multiplying,

= $\sf{AB = 7.2 \times 2}$

$\sf{\implies AB = 14.4\:m}$

$\boxed{\therefore \underline{ \sf{The\:length\:of\:the\:ladder\:is\:14.4\:m}}}$

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Additional Information:-

• Trigonometric ratios are only applicable for right - angled triangle.

The trigonometric table is as follows:-

$\boxed{\begin{array}{c|c|c|c|c|c} \theta & \sf{0^{\circ}} & \sf{30^{\circ}} & \sf{45^{\circ}} & \sf{60^{\circ}} & \sf{90^{\circ}} \\ \dfrac{\qquad \qquad}{} & \dfrac{\qquad \qquad}{} & \dfrac{\qquad \qquad}{} & \dfrac{\qquad \qquad}{} & \dfrac{\qquad\qquad}{} & \dfrac{\qquad \qquad}{} \\ \sf{Sin\theta} & \sf{0} & \sf{\dfrac{1}{2}} & \sf{\dfrac{1}{\sqrt{2}}} & \sf{\dfrac{\sqrt{3}}{2}} & \sf{1} \\ \sf{Cos\theta} & \sf{1} & \sf{\dfrac{\sqrt{3}}{2}} & \sf{\dfrac{1}{\sqrt{2}}} & \sf{\dfrac{1}{2}} & \sf{0} \\ \sf{Tan\theta} & \sf{0} & \sf{\dfrac{1}{\sqrt{3}}} & \sf{1} & \sf{\sqrt{3}} & \sf{Not\:defined} \\ \sf{Cosec\theta} & \sf{Not\:Defined} & \sf{2} & \sf{\sqrt{2}} & \sf{\dfrac{2}{\sqrt{3}}} & \sf{1} \\ \sf{Sec\theta} & \sf{1} & \sf{\dfrac{2}{\sqrt{3}}} & \sf{\sqrt{2}} & \sf{2} & \sf{Not\:Defined} \\ \sf{Cot\theta} & \sf{Not\:Defined} & \sf{\sqrt{3}} & \sf{1} & \sf{\dfrac{1}{\sqrt{3}}} & \sf{0} \end{array}}$

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Some other identities to be known.

In a right-angled triangle ABC, right angled at B (as shown in figure):-

• SinC = $\sf{\dfrac{Perpendicular}{Hypotenuse} = \dfrac{AB}{AC}}$

• CosC = $\sf{\dfrac{Base}{Hypotenuse} = \dfrac{BC}{AC}}$

• TanC = $\sf{\dfrac{Perpendicular}{Base} = \dfrac{AB}{BC}}$

• CosecC = $\sf{\dfrac{Hypotenuse}{Perpendicular} = \dfrac{AC}{AB}}$

• SecC = $\sf{\dfrac{Hypotenuse}{Base} = \dfrac{AC}{BC}}$

• CotC = $\sf{\dfrac{Base}{Perpendicular} = \dfrac{BC}{AB}}$

Here we took ratio as C because from the attached figure we can see the. reference angle is C.

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