Ex.8.1 Q6 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Question

If \(\,\,\angle {A}\) and \(\,\,\angle {B}\) are acute angles such that \(\,\,\text{cos}\,{A}=\text{cos}\,{B},\) then show that \(\angle {A}=\angle \,{B}\,{.}\)

Text Solution

What is the known?

\(\angle {A}\) and \(\,\,\angle {B}\) are acute angles and \(\begin{align}\text{cos}\, A  = \text{cos}\,  B\end{align}\)

What is the unknown?

To show that \(\angle A= \angle B\)

Reasoning:

Using \(\text{cos} \,A\) and \(\text{cos}\, B,\) we can find the ratio of the length of two sides of the right-angled triangle with respective angles. Then compare both the ratios.

Steps:

In the right-angled triangle \(ABC, \,\)\(\,\angle A\)  and  \(\angle B \) are acute angles and \(\angle C\) is right angle.

\[\begin{align} & \text{cos}\,{A}\,{=}\,\frac{\text{side adjacent to}\ \angle {A}}{\text{hypotenuse}}{=}\frac{{AC}}{{AB}} \\ & \text{cos}\,{B}\,{=}\,\frac{\text{side adjacent to}\,\angle {B}}{\text{hypotenuse}}{=}\,\frac{{BC}}{{AB}} \end{align}\]

Given that \(\begin{align} \;\text{cos} \,A&=\,\text{cos} B \end{align}\)

\(\begin{align} \text{Therefore, }\frac{{AC}}{{AB}}\,&=\,\frac{{BC}}{{AB}} \\ {AC}\,&=\,{BC}\end{align}\)

Hence, \(\angle A=\angle B\,,\) (angles opposite to equal sides of triangle are equal)

Alternatively,

Reasoning:

Using \(\text{ cos} \,A\) and \(\text{cos}\, B,\) we can find the ratio of the length of two sides of the right-angled triangle with respective angles. Then by using Pythagoras theorem, relation between the sides

Let us consider a triangle \(ABC\) in which \(\begin{align}{CO} \perp {AB}\end{align}\)

It is given that

\[\begin{align} \text{cos}\,{A}\,&=\, \text{cos}\,{B} \\ \frac{{AO}}{{AC}}&=\frac{{BO}}{{BC}} \\ \frac{{AO}}{{BO}}\,&=\,\frac{{AC}}{{BC}}\\ \\ \text{ Let }\; \frac{ {A O}}{{B O}}&=\frac{{A C}}{{B C}}={k} \\ { A O}&={k B O}\;\dots\rm{(i)} \\ {A C}&={k B C}\;\dots \rm{(ii)} \end{align}\]

By applying Pythagoras theorem in  \(\Delta {CAO}\) and \(\Delta {CBO},\) we get.

\[\begin{align}  {A}{{{C}}^{{2}}} & ={A}{{{O}}^{{2}}}{+C}{{{O}}^{{2}}}[\text{from} \;\Delta\;{ CAO}] \\  {C}{{{O}}^{{2}}}&={A}{{{C}}^{{2}}}-{A}{{{O}}^{{2}}}\cdots \left(\rm {iii} \right) \\  {B}{{{C}}^{{2}}}&={B}{{{D}}^{{2}}}{+C}{{{O}}^{{2}}}[\text {from}\;\Delta\,{ CBO}] \\  {C}{{{O}}^{{2}}}&={B}{{{C}}^{{2}}}\,-{B}{{{O}}^{{2}}}\cdots \left( \rm{iv} \right) \\ \end{align}\]

From equation (iii) and equation (iv), we get

\[\begin{align} {A}{{{C}}^{{2}}}-{A}{{{O}}^{{2}}}&=\,{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}} \\ \ {{{(kBC)}}^{{2}}}-{{{(kBO)}}^{{2}}}&=\,{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}} \\  {{k}^{{2}}}{B}{{{C}}^{{2}}}-{{k}^{{2}}}{B}{{{O}}^{{2}}}\,&=\,{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}} \\  {{k}^{{2}}}{(B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}}{)}\,&=\,{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}} \\  {{k}^{{2}}}\,&=\,\frac{{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}}}{{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}}} \\ &=1 \\k\,&=1 \end{align}\]

Putting this value in equation (ii) we obtain

\[{AC}\,{=}\,{BC}\]

\({\angle A=\angle B}\) (angles opposite to equal sides of triangle are equal.)