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.
ExampleThe 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
Example: in our ladder example we know the length of:
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:
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 ... Example:
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. Step By StepThese are the four steps we need to follow:
ExamplesLet’s look at a couple more examples:
ExampleFind the angle of elevation of the plane from point A on the ground.
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.
ExampleFind the size of angle a°
cos a° = 6,750/8,100 = 0.8333 cos-1 of 0.8333 = 33.6° (to 1 decimal place) 250, 1500, 1501, 1502, 251, 1503, 2349, 2350, 2351, 3934 Copyright © 2021 MathsIsFun.com |