Question

1. if a, b and c are the angles of a triangle with b + c = 90°, which of the following are true? i. tan b tan c = sin a ii. sin(90° - a)+sin(90° - b)=sin c iii. cos²a = cos²b + cos²c a. i and ii only b. i and iii only c. ii and iii only d. i, ii and iii 15. if x, y and z are the angles of a triangle with y + z = 90° and y < z, which of the following are true? i. tan(x + y)<tan z ii. sin(90° - y)>sin(90° - z)

Explanation:

Step1: Recall trigonometric identities

We know that $\sin(90^{\circ}-\alpha)=\cos\alpha$ and $\tan(90^{\circ}-\alpha)=\cot\alpha$. Also, for angles $A,B,C$ of a triangle, $A + B + C=180^{\circ}$. Given $B + C = 90^{\circ}$, then $A = 90^{\circ}$.

Step2: Analyze statement I

If $\tan B\tan C=\sin A$, since $A = 90^{\circ}$, $\sin A = 1$. And in a right - triangle with $B + C=90^{\circ}$, $\tan B=\frac{\sin B}{\cos B}$ and $\tan C=\frac{\sin C}{\cos C}$. Since $C = 90^{\circ}-B$, $\tan B\tan(90^{\circ}-B)=\tan B\cot B = 1=\sin A$. So statement I is true.

Step3: Analyze statement II

$\sin(90^{\circ}-A)+\sin(90^{\circ}-B)=\sin C$. Since $A = 90^{\circ}$, $\sin(90^{\circ}-A)=0$. And since $B + C=90^{\circ}$, $\sin(90^{\circ}-B)=\cos B=\sin C$. So $\sin(90^{\circ}-A)+\sin(90^{\circ}-B)=\sin C$ is true.

Step4: Analyze statement III

$\cos^{2}A+\cos^{2}B+\cos^{2}C$. Since $A = 90^{\circ}$, $\cos A=0$. And since $C = 90^{\circ}-B$, $\cos C=\sin B$. Then $\cos^{2}A+\cos^{2}B+\cos^{2}C=0+\cos^{2}B+\sin^{2}B = 1$. So statement III is true.

For the second problem:
Given $y + z=90^{\circ}$, so $x = 90^{\circ}$.
$\tan(x + y)<\tan z$, since $x = 90^{\circ}$, $\tan(x + y)=\tan(90^{\circ}+y)=-\cot y$. And $z>y$ and $y + z = 90^{\circ}$. $\tan z>\tan y>0$, $-\cot y<0$. So $\tan(x + y)<\tan z$.
$\sin(90^{\circ}-y)>\sin(90^{\circ}-z)$, since $y\cos z$.

All statements in both problems are true.

Answer:

1. D. I, II and III
2. D. I, II and III