Andhra Pradesh SSC (10th class) Mathematics Trigonometry Exercise - 11.1 Solutions -part-1

Mathematics trigonometry Exercise - 11.1

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Mathematics Trigonometry Exercise

6. If ∠a and ∠x are acute angles   such that Cos A = Cos X then such that   ∠a = ∠x

Sol: Given that   Cos A = Cos X   ------   (1)                

We have   Cos A =AB/AC and   Cos X = XY /XY 

AB/AC = XY/XZ   From  Eg (1)                

AB/AC = XY/XZ = k ------   (2)

From ⧍ABC 

BC ² =AC ²-AB ²

BC ² = AC ²   - (k AC) ²

BC ²=AC ²   -   k ² AC ²

BC ²=AC ²   (1- k ²)

BC =√ AC ²     (1- k ²)

BC =√ AC ²   (1- k ²)

BC = AC √ (1- k ²)

 From ⧍ XYZ

YZ ² =AC ²-XY ²

YZ ² = XZ ²   - (k XZ) ²

YZ ²= XZ ²   -   k ² XZ ²

YZ ²= XZ ²   (1- k ²)

YZ =√ XZ ²   (1- k ²)

YZ = XZ   √ (1- k ²)

BC/ YZ = AC √ (1- k ²)/ XZ   √ (1- k ²)

BC/ YZ = AC/ XZ   ------   (3)

From (2) & (3) Equations

Hence   AB/XY= BC/ YZ = AC/ XZ   

⧍ ABC ~ ⧍ XYZ (   .  ̇. ∠A =∠X   )

7. Cot 𝜽 =7/8, evaluate (1)   (1+sin 𝜽) (1-sin 𝜽) / (1+cos 𝜽) (1-cos 𝜽)   (2) (1+sin 𝜽)/ Cos 𝜽

Sol:   Cot 𝜽 =7/8   

From ⧍ABC                          

AB ²= BC ²+AC ²

  AB ²= 7²+8²

AB ²= 49+64

AB ²=113

AB =√113

2)(1+sin 𝜽) (1-sin 𝜽) /(1+cos 𝜽) (1-cos 𝜽)   

 = 1-sin ² 𝜽/1-cos ² 𝜽

= cos ² 𝜽 /sin ² 𝜽

= Cot ² 𝜽 

= (BC/AB) ²

= (7/√113) ²

= 7²/(√113) ²

=49/113

(2)(1+sin 𝜽)/ Cos 𝜽

=1/Cos 𝜽+Sin 𝜽/Cos 𝜽

=Sec 𝜽+Tan 𝜽

=AB/BC+AC/BC

=√113/ 7+8/7

=√113+8/7 

8. In a Right angle triangle ABC, Right angle is at B, Tan A =√3 then find   the value of   1) Sin A Cos C + Cos A Sin C  2) Cos A Cos C - Sin A Sin C

Sol:    Given that   Tan A   = √3/1

(Hypotenuse) ²= (side) ²+ (side) ²

(Hypotenuse) ²= (√3) ² +1

=3+1

=4

1 1) Sin A Cos C + Cos A Sin C       

=√3/2 ×√3/2+1/2×1/2

=3/4 +1/4

=4/4

=1

2) Cos A Cos C - Sin A Sin C

=1/2×√3/2 - 1/2×√3/2 

=√3/4 -√3/4

=0