Question

Solve the ques 15 in the attachment

Class 10

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Answer 1

hii!!!

here's ur answer..

consider angle b = 90° in ∆abc

given the value of tan A = 1/√3

we know that tan A = perpendicular/base

therefore in ∆abc , perpendicular = 1 and base = √3

by Pythagoras theorem :-

hypotenuse = √ [ (1)² + (√3)² ]

= √ ( 1 + 3 )

= √4

= 2

hence, the hypotenuse in ∆abc is 2.

for angle A, we have base = √3, perpendicular = 1 and hypotenuse = 2

sin A = perpendicular/hypotenuse = 1/2

and cos A = base/hypotenuse = √3/2

for angle B, we have base = 1, perpendicular = √3 and hypotenuse = 2

Sin B = perpendicular/hypotenuse = √3/2

cos B = base/hypotenuse = 1/2

sin A cos B + cos A sin B

= 1/2 × 1/2 + √3/2 × √3/2

= 1/4 + 3/4

= 4/4

= 1

hope this helps..!!

Answer 2

$Given : tanA = \frac{1}{\sqrt{3}}$

= > A = 30.

Given right angled triangle at C. So, C = 90.

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Now,

We know that sum of interior angles of a triangle = 180.

= > A + B + C = 180

= > 30 + B + 90 = 180

= > 120 + B = 180

= > B = 60.

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Now,

sinA cosB + cosA sinB = sin(A + B)

= sin(30 + 60)

= sin(90)

= 1.

Therefore, the final result is 1.

Hope this helps!