## Friday, October 2, 2020

• October 02, 2020
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• 1 comment

## Mathematics trigonometry Exercise - 11.1

Hi friends and my dear students! In this post, I have covered Andhra Pradesh SSC(10th class) Mathematics trigonometry Exercise - 11.1 Solutions part-1. After Reading Mathematics trigonometry Exercise, Please do share it with your friends. You can 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 272+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

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