October 02, 2020
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)
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
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