Pls ans both the questions im very confused

Mathematics

1 answer:

lions [1.4K]3 years ago

5 0

Answer:

Step-by-step explanation:

In Δ PQS and ΔPQT

∠QPS = ∠RPT {Common angle}

∠PSQ = ∠PTR = 90° {QS & RT are altitude}

QS = RT {given}

ΔPQS ≅ ΔPRT {A A S congruent}

PQ = PS {CPCT}

So, PQR is an isosceles triangle.

a) Flas card (I) and (iii) are congruent.

b) A S A congruent

c) BC ≅ RP

You might be interested in

A company's stock was selling at

Mandarinka [93]

Answer:

10%

Step-by-step explanation:

5 0

3 years ago

1. Make up two equations, one that is true and one that is false. Do not state which equation is true and which is false. Your c

N76 [4]

1. The equations that I can think of are,
3 + 1 = 11 - 6
and
8 - 1 = 6 + 1

2. An example of equation that uses the variable x,
6x - 2x = 8
It may be required of you to perform certain steps in order to determine the value of x.

6 0

3 years ago

Read 2 more answers

If sin(x) = 5/13, and x is in quadrant 1, then tan(x/2) equals what?

Rufina [12.5K]

$x$ is in quadrant I, so $0$ , which means $0$ , so $\dfrac x2$ belongs to the same quadrant.

Now,

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

Since $\sin x=\dfrac5{13}$ , it follows that

$\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}$

Since $x$ belongs to the first quadrant, you take the positive root ( $\cos x>0$ for $x$ in quadrant I). Then

$\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}$

$\tan x$ is also positive for $x$ in quadrant I, so you take the positive root again. You're left with

$\tan\dfrac x2=\dfrac15$