Question

In the figure below, it is given that angle C = 90°, AD = DB, ED is perpendicular to AB, AB = 20 units and
AC = 12 units.
7. Area of triangle AEC is
(A) 24 sq. units (B) 21 sq. units
(C) 42 sq. units (D) 21/2 sq.units
8. The value of tan (delta+beta) is
(A) – 117/44
(B) 17/4
(C) 3/4
(D) 5/4
9. The value of cos (alpha+beta) is
(A) 4/5
(B) 3/5
(C) 117/125
(D) –44/125

430a81a81b569f5c34702e238025294c.docx

Answer 1

Answer:

4/5 please follow me please

Answer 2

Answer:

(7) 21 sq. unit,  (8) $-\frac{117}{44}$   (9) $\frac{4}{5}$

Step-by-step explanation:

(7)

$\mathrm{AD}=\mathrm{DB}$

$\angle \mathrm{ADE}=\angle \mathrm{BDE}$

$\mathrm{ED}=\mathrm{ED}$(Common to both triangles)

Hence both $\Delta$ 's are congruent

So, $\mathrm{AE}=\mathrm{BE}$

$\beta=\gamma$

$\mathrm{AC}=12, \mathrm{AB}=20, \mathrm{BC}=16$

$\tan \gamma=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{12}{16}=\frac{3}{4}$

$\alpha+\beta=90-\gamma$

$\alpha+\gamma=90-\gamma$

$\alpha=90-2 \gamma$

$\tan \alpha=\frac{\mathrm{CE}}{\mathrm{AC}}$

$\mathrm{CE}=\mathrm{AC} \tan (90-2 \gamma)$

$\mathrm{CE}=\mathrm{AC} \tan (90-2 \gamma)=12 \times \frac{\left(1-\tan ^{2} \gamma\right)}{2 \tan _{\gamma}}$

                                   $=12 \times \frac{\left(1-\frac{9}{16}\right)}{2 \times \frac{3}{4}}=8 \times \frac{7}{16}=\frac{7}{2}$

$A C=12, A B=20, B C=16$

$\tan \gamma=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{12}{16}=\frac{3}{4}$

$\alpha+\beta=90-\gamma$

$\alpha+\gamma=90-\gamma$

$\alpha=90-2 \gamma$

$\tan \alpha=\frac{\mathrm{CE}}{\mathrm{AC}}$

$\mathrm{CE}=\mathrm{AC} \tan (90-2 \gamma)$

$\mathrm{CE}=\mathrm{AC} \tan (90-2 \gamma)=12 \times \frac{\left(1-\tan ^{2} \gamma\right)}{2 \tan \gamma}$

                                    $=12 \times \frac{\left(1-\frac{9}{16}\right)}{2 \times \frac{3}{4}}=8 \times \frac{7}{16}=\frac{7}{2}$

Area $\frac{1}{2} \times \mathrm{AC} \times \mathrm{CE}$$=\frac{1}{2}\times 12 \times \frac{7}{2}=21 \mathrm{sq} . \text { units }$

Hence option B is correct.

(8)

$\begin{aligned}&\delta=\gamma+\beta \\&\delta+\beta=3 \gamma \\&\tan (3 \gamma) \\&=\frac{3 \tan \gamma-\tan ^{3} \gamma}{1-3 \tan ^{2} \gamma}=-\frac{117}{44}\end{aligned}$

Hence option A is correct.

(9)

$cos(\alpha +\beta )=\frac{4}{5}$

#SPJ2