Last updated at May 29, 2018 by Teachoo
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Ex7.1, 8 In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that: ΔAMC ≅ ΔBMD Given: ∠ ACB = 90° M is the mid-point of AB So, AM = BM Also, DM = CM To prove: ΔAMC ≅ ΔBMD Proof: Lines CD & AB intersect… So, ∠AMC = ∠BMD In ΔAMC and ΔBMD, AM = BM ∠AMC = ∠BMD CM = DM ∴ ΔAMC ≅ ΔBMD Ex7.1, 8 In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that: (ii) ∠DBC is a right angle. From part 1, ΔAMC ≅ ΔBMD ∴ ∠ACM = ∠BDM But ∠ACM and ∠BDM are alternate interior angles for lines AC & BD If a transversal intersects two lines such that pair of alternate interior angles is equal, then lines are parallel. So, BD || AC Now, Since DB || AC and Considering BC as transversal , ⇒ ∠DBC + ∠ACB = 180° ⇒ ∠DBC + 90° = 180° ⇒ ∠DBC = 180° – 90° ⇒ ∠DBC = 90° Hence, ∠ DBC is a right angle Hence proved Ex 7.1, 8 In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that: (iii) ΔDBC ≅ ΔACB From part 1, ΔAMC ≅ ΔBMD AC = BD In ΔDBC and ΔACB, DB = AC ∠DBC = ∠ACB BC = CB ∴ ΔDBC ≅ ΔACB Ex7.1, 8 In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that: (iv) CM = 1/2 AB From part 3, ΔDBC ≅ ΔACB ∴ DC = AB 1/2 DC = 1/2 AB CM = 1/2 AB Hence proved
We can find an unknown angle in a right-angled triangle, as long as we know the lengths of two of its sides.
Example
The ladder leans against a wall as shown.
What is the angle between the ladder and the wall?
The answer is to use Sine, Cosine or Tangent!
But which one to use? We have a special phrase "SOHCAHTOA" to help us, and we use it like this:
Step 1: find the names of the two sides we know
- Adjacent is adjacent to the angle,
- Opposite is opposite the angle,
- and the longest side is the Hypotenuse.
Example: in our ladder example we know the length of:
- the side Opposite the angle "x", which is 2.5
- the longest side, called the Hypotenuse, which is 5
Step 2: now use the first letters of those two sides (Opposite and Hypotenuse) and the phrase "SOHCAHTOA" to find which one of Sine, Cosine or Tangent to use:
SOH... | Sine: sin(θ) = Opposite / Hypotenuse |
...CAH... | Cosine: cos(θ) = Adjacent / Hypotenuse |
...TOA | Tangent: tan(θ) = Opposite / Adjacent |
In our example that is Opposite and Hypotenuse, and that gives us “SOHcahtoa”, which tells us we need to use Sine.
Step 3: Put our values into the Sine equation:
Sin (x) = Opposite / Hypotenuse = 2.5 / 5 = 0.5
Step 4: Now solve that equation!
sin(x) = 0.5
Next (trust me for the moment) we can re-arrange that into this:
x = sin-1(0.5)
And then get our calculator, key in 0.5 and use the sin-1 button to get the answer:
x = 30°
And we have our answer!
Well, the Sine function "sin" takes an angle and gives us the ratio "opposite/hypotenuse",
But sin-1 (called "inverse sine") goes the other way ...
... it takes the ratio "opposite/hypotenuse" and gives us an angle.
Example:
- Sine Function: sin(30°) = 0.5
- Inverse Sine Function: sin-1(0.5) = 30°
| On the calculator press one of the following (depending on your brand of calculator): either '2ndF sin' or 'shift sin'. |
On your calculator, try using sin and sin-1 to see what results you get!
Also try cos and cos-1. And tan and tan-1.
Go on, have a try now.
Step By Step
These are the four steps we need to follow:
- Step 1 Find which two sides we know – out of Opposite, Adjacent and Hypotenuse.
- Step 2 Use SOHCAHTOA to decide which one of Sine, Cosine or Tangent to use in this question.
- Step 3 For Sine calculate Opposite/Hypotenuse, for Cosine calculate Adjacent/Hypotenuse or for Tangent calculate Opposite/Adjacent.
- Step 4 Find the angle from your calculator, using one of sin-1, cos-1 or tan-1
Examples
Let’s look at a couple more examples:
Example
Find the angle of elevation of the plane from point A on the ground.
- Step 1 The two sides we know are Opposite (300) and Adjacent (400).
- Step 2 SOHCAHTOA tells us we must use Tangent.
- Step 3 Calculate Opposite/Adjacent = 300/400 = 0.75
- Step 4 Find the angle from your calculator using tan-1
Tan x° = opposite/adjacent = 300/400 = 0.75
tan-1 of 0.75 = 36.9° (correct to 1 decimal place)
Unless you’re told otherwise, angles are usually rounded to one place of decimals.
Example
Find the size of angle a°
- Step 1 The two sides we know are Adjacent (6,750) and Hypotenuse (8,100).
- Step 2 SOHCAHTOA tells us we must use Cosine.
- Step 3 Calculate Adjacent / Hypotenuse = 6,750/8,100 = 0.8333
- Step 4 Find the angle from your calculator using cos-1 of 0.8333:
cos a° = 6,750/8,100 = 0.8333
cos-1 of 0.8333 = 33.6° (to 1 decimal place)
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