Answer to Question #232659 in Trigonometry for dizzo

Question

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

Answer 1

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

Consider right triangle $BAC$

$$\tan \theta=\dfrac{opposite}{adjacent}=\dfrac{BC}{AC}=\dfrac{56 m}{113 m}$$

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

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

Then $\alpha=\theta=26.36\degree.$

(b)

$$6x^2+ x +7=6x^2-6xm+6xm+x$$

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

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

$$+6m^2+m+7$$

$$6x^2+ x +7=6x^2+12xm-12xm+x$$

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

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

$$+24m^2-2m+7$$

The remainder is the same

$$6m^2+m+7=24m^2-2m+7$$

$$18m^2-3m=0$$

$$3m(6m-1)=0$$

$$m_1=0, m_2=\dfrac{1}{6}$$