Answer to Question #232659 in Trigonometry for dizzo

Question #232659

Find the angle of elevation to the top of a 56 m high building from point A which is 113 m

from its base. What is the angle of depression from the top of the building to A? /6mks

(b) The expression 6x^2+ x +7 leaves the same remainder when divided by x – m and by x+2m,


where m ≠ 0. Calculate the value of m./4mks


1
Expert's answer
2021-09-06T13:08:22-0400

(a) Let θ=\theta= the angle of elevation to the top of a building from point A.A.

Consider right triangle BACBAC


tanθ=oppositeadjacent=BCAC=56m113m\tan \theta=\dfrac{opposite}{adjacent}=\dfrac{BC}{AC}=\dfrac{56 m}{113 m}

θ=tan15611326.36°\theta=\tan^{-1}\dfrac{56}{113}\approx26.36\degree

Let α=\alpha= the angle of depression from the top of the building to A.A.

Then α=θ=26.36°.\alpha=\theta=26.36\degree.


(b)

6x2+x+7=6x26xm+6xm+x6x^2+ x +7=6x^2-6xm+6xm+x

(6m+1)(m)+(6m+1)(m)+7-(6m+1)(m)+(6m+1)(m)+7

=6x(xm)+(6m+1)(xm)=6x(x-m)+(6m+1)(x-m)

+6m2+m+7+6m^2+m+7



6x2+x+7=6x2+12xm12xm+x6x^2+ x +7=6x^2+12xm-12xm+x

(12m1)(2m)+(12m1)(2m)+7-(12m-1)(2m)+(12m-1)(2m)+7

=6x(x+2m)(12m1)(x+2m)=6x(x+2m)-(12m-1)(x+2m)

+24m22m+7+24m^2-2m+7

The remainder is the same


6m2+m+7=24m22m+76m^2+m+7=24m^2-2m+7

18m23m=018m^2-3m=0

3m(6m1)=03m(6m-1)=0

m1=0,m2=16m_1=0, m_2=\dfrac{1}{6}


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